Theodore Norvell (C) 1999 with updates later on.
This article is about parsing expressions such as a*b  a*d  e*f using a technique known as recursive descent. I've assumed you know at least a little bit about contextfree grammars and parsing.
Parsing expressions by recursive descent poses two classic problems
The classic solution to the first problem does not solve the second. I will present the classic solution, a well known alternative known as the "Shunting Yard Algorithm", and a less well known one that I have called "Precedence Climbing".
Consider the following example grammar, G,
E > E "+" E  E "" E  "" E  E "*" E  E "/" E  E "^" E  "(" E ")"  v
in which v is a terminal representing identifiers and/or constants.
We want to build a parser that will
Each input in the language will have a single AST based on the following precedence and associativity rules:
For example the first three rules tell us that
a ^ b * c ^ d + e ^ f / g ^ (h + i)
parses to the tree
+( *( ^(a,b), ^(c,d) ), /( ^(e,f), ^(g,+(h,i)) ) )
while the last rule tells us that
a  b  c
parses to ((a,b),c) rather than (a,(b,c)), whereas
a ^ b ^ c
parses to ^(a, ^(b,c)) rather than ^(^(a,b), c).
The precedence of binary ^ over unary  tells us that
 a ^  b
parses to (^(a, (b))). Some programming language designers choose to put unary operators at the highest level of precedence. I chose to give unary  a lower precedence than *, /, and ^ because having some binary operators with higher precedence than some unary operators makes the parsing problem just a bit more challenging and raises some issues that otherwise wouldn't come up.
Aside: I am assuming that the desired output of the parser is an abstract syntax tree (AST). The same considerations arise if the output is to be some other form such as reversepolish notation (RPN), calls to an analyzer and code generator (for onepass compilers), or a numerical result (as in a calculator). All the algorithms I present are easily modified for these forms of output.
The idea of recursivedescent parsing is to transform each nonterminal of a grammar into a subroutine that will recognize exactly that nonterminal in the input.
Left recursive grammars, such as G, are unsuitable for recursivedescent parsing because a leftrecursive production leads to an infinite recursion. While the parser may be partially correct, it may not terminate.
We can transform G to an equivalent nonleftrecursive grammar G1 as follows:
E > P {B P} P > v  "(" E ")"  U P B > "+"  ""  "*"  "/"  "^" U > ""
The braces "{" and "}" represent zero or more repetitions of what is inside of them. Thus you can think of E as having an infinity of alternatives:
E > P  P B P  P B P B P  ... ad infinitum
The language described by this grammar is the same as that of grammar G, that is, L(G1) = L(G).
Not only is left recursion eliminated, but the G1 is unambiguous and each choice can be made by looking at the next token in the input.
Aside: Technically, G1 is an example of what is called an LL(1) grammar. I don't want to make this essay more technical than it needs to be, so I'm not going to stop and go into what that means. End of Aside.
Let's look at a recursive descent recognizer based on this grammar. I call this algorithm a recognizer rather than a parser because all it does is to recognize whether the input is in the language of the grammar or not. It does not produce an abstract syntax tree, or any other form of output that represents the contents of the input.
I'll assume that the following subroutines exist:
Using these, let's construct a subroutine "expect", which I will use throughout this essay
expect( tok ) is if next = tok consume else error
We will now write a subroutine called "Erecognizer". If it does not call "error", then the input was an expression according to the above grammars. If it does call "error", then the input contained a syntax error, e.g., unmatched parentheses, a missing operator or operand, etc.
Erecognizer is E() expect( end )
E is P while next is a binary operator consume P
P is if next is a v consume else if next = "(" consume E expect( ")" ) else if next is a unary operator consume P else error
Notice how the structure of the recognition algorithm mirrors the structure of the grammar. This is the essence of recursive descent parsing.
The difference between a recognizer and a parser is that a parser produces some kind of output that reflects the structure of the input. Next we will look at a way to modify the above recognition algorithm to be a parsing algorithm. It will build an AST, according to the precedence and associativity rules, using a method known as the "shunting yard" algorithm.
The idea of the shunting yard algorithm is to keep operators on a stack until both their operands have been parsed. The operands are kept on a second stack. The shunting yard algorithm can be used to directly evaluate expressions as they are parsed (it is commonly used in electronic calculators for this task), to create a reverse Polish notation translation of an infix expression, or to create an abstract syntax tree. I'll create an abstract syntax tree, so my operand stacks will contain trees.
The key to the algorithm is to keep the operators on the operator stack ordered by precedence (lowest at bottom and highest at top), at least in the absence of parentheses. Before pushing an operator onto the operator stack, all higher precedence operators are cleared from the stack. Clearing an operator consists of removing the operator from the operator stack and its operand(s) from the operand stack, making a new tree, and pushing that tree onto the operand stack. At the end of an expression the remaining operators are put into trees with their operands and that is that.
The following table illustrates the process for an input of x*y+z. Stacks are written with their tops to the left. The sentinel value acts as an operator of lowest precedence.
Remaining input  Operand Stack  Operator Stack  Next Action  
x * y + z end  sentinel  Push x on to the operand stack.  
* y + z end  x  sentinel  Compare the precedence of * with the precedence of the sentinel.  
* y + z end  x  sentinel  It's higher, so push * on to the operator stack  
y + z end  x  binary(*) sentinel  Push y on to the operand stack.  
+ z end  y x  binary(*) sentinel  Compare the precedence of + with the precedence of * .  
+ z end  y x  binary(*) sentinel  It's lower, so make a tree from *, y, and x .  
+ z end  *(x,y)  sentinel  Compare the precedence of + with the precedence of the sentinel.  
+ z end  *(x,y)  sentinel  It's higher, so push + on to the operator stack.  
z end  *(x,y)  binary(+) sentinel  Push z on to the operator stack .  
end  z *(x,y)  binary(+) sentinel  Make a tree from +, z, and *(x,y).  
end  +( *(x,y), z )  sentinel 
Compare this to parsing x + y * z.
Remaining input  Operand Stack  Operator Stack  Next Action  
x + y * z end  sentinel  Push x on to the operand stack  
+ y * z end  x  sentinel  Compare the precedence of + with the precedence of the sentinel  
+ y * z end  x  sentinel  It's higher, so push + on to the operator stack  
y * z end  x  binary(+) sentinel  Push y on to the operand stack  
* z end  y x  binary(+) sentinel  Compare the precedence of * with the precedence of +.  
* z end  y x  binary(+) sentinel  It's higher so, push * on to the operator stack  
z end  y x  binary(*) binary(+) sentinel  Push z on to the operand stack  
end  z y x  binary(*) binary(+) sentinel  Make a tree from *, y, and z  
end  *(y, z) x  binary(+) sentinel  Make a tree from +, x, and *(y,z)  
end  +( x, *(y, z) )  sentinel 
In addition to "next", "consume", "end", "error", and "expect", which are explained in the previous section, I will assume that the following subroutines and constants exist:
In the algorithm that follows, I compare operators and the sentinel with a > sign. This comparison is defined as follows:
Now we define the following subroutines:
Aside: I hope the pseudocode notation is fairly clear. I'll just comment that I'm assuming that parameters are passed by reference, so only 2 stacks are created throughout the execution of EParser.
Eparser is var operators : Stack of Operator := empty var operands : Stack of Tree := empty push( operators, sentinel ) E( operators, operands ) expect( end ) return top( operands )
E( operators, operands ) is P( operators, operands ) while next is a binary operator pushOperator( binary(next), operators, operands ) consume P( operators, operands ) while top(operators) not= sentinel popOperator( operators, operands )
P( operators, operands ) is if next is a v push( operands, mkLeaf( v ) ) consume else if next = "(" consume push( operators, sentinel ) E( operators, operands ) expect( ")" ) pop( operators ) else if next is a unary operator pushOperator( unary(next), operators, operands ) consume P( operators, operands ) else error
popOperator( operators, operands ) is if top(operators) is binary const t1 := pop( operands ) const t0 := pop( operands ) push( operands, mkNode( pop(operators), t0, t1 ) ) else push( operands, mkNode( pop(operators), pop(operands) ) )
pushOperator( op, operators, operands ) is while top(operators) > op popOperator( operators, operands ) push( op, operators )
Usually the shunting yard algorithm is presented without the use of recursion. This may be more efficient and might aid in generating better error messages, but I find the code a bit harder to understand.
The classic solution to recursivedescent parsing of expressions is to create a new nonterminal for each level of precedence as follows. G2:
E > T {( "+"  "" ) T} T > F {( "*"  "/" ) F} F > P ["^" F] P > v  "(" E ")"  "" T
(The brackets [ and ] enclose an optional part of the production. As before, the braces { and } enclose parts of the productions that may be repeated 0 or more times, and  separates alternatives. The unquoted parentheses ( and ) serve only to group elements in a production.)
Grammar G2 describes the same language as the previous two grammars: L(G2) = L(G1) = L(G)
The grammar is ambiguous; for example, x*y has two parse trees: E(T(F(P("", T(F(P("x")), "*"))))) and E(T(F(P("",T(F(P("x"))))),"*",F(P("y")))). The ambiguity is resolved by staying in each loop (in the productions for E and T) as long as possible and by taking the option —if possible— in the production for F. With these policy in place, all choices can be made by looking only at the next token of input.
Aside: If our precedence had been such that our unary operator had highest precedence, then the grammar would not have been ambiguous. For those who are interested in such things, I'll note that this grammar is not LL(1); LL(1) grammars are never ambiguous. Nevertheless everything works out just fine, if we adopt the policies mentioned in the previous paragraph. End of Aside.
Note that the leftassociative and the rightassociative operators are treated differently; leftassociative operators are consumed in a loop, while rightassociative operators are handled with rightrecursive productions. This is to make the tree building a bit easier.
Here is an example of parsing a*b  c^d  e*f by recursive descent.
Each contour line shows what is recognized by each invocation of E, T, or F. For instance we can see that the top level call to E invokes T three times; these three invocations of T respectively recognize a*b, c^d, and e*f. Not shown are the calls to P, of which there is one for each variable. Another way to look at it is that the contour lines show the parse tree (or would if I'd included the contour lines for P). The solid lines show the AST that we would like to be constructed.
We can transform this grammar to a parser written in pseudo code.
Eparser is var t : Tree t := E expect( end ) return t
E is var t : Tree t := T while next = "+" or next = "" const op := binary(next) consume const t1 := T t := mkNode( op, t, t1 ) return t
T is var t : Tree t := F while next = "*" or next = "/" const op := binary(next) consume const t1 := F t := mkNode( op, t, t1 ) return t
F is var t : Tree t := P if next = "^" consume const t1 := F return mkNode( binary("^"), t, t1) else return t
P is var t : Tree if next is a v t := mkLeaf( next ) consume return t else if next = "(" consume t := E expect( ")" ) return t else if next = "" consume t := F return mkNode( unary(""), t) else error
It may be worthwhile to trace this algorithm on a few example inputs.
This classic solution has a few drawbacks:
When there are a large number of precedence levels, as in the C and C++ languages, the first two disadvantages become problematic. In Pascal the number of precedence levels was deliberately kept small because —I suspect— its designer, Niklaus Wirth, was aware of the shortcomings of this method when the number of precedence levels is large.
The size problem can be overcome by creating one subroutine that is parameterized by precedence level rather than writing a separate routine for each level. But the speed problem remains. Note that the number of calls to parse an expression consisting of a single identifier is proportional to the number of levels of precedence.
For languages in which the set of operators and their precedences and associativity are not hardcoded, we need a more flexible approach.
A method that solves all the listed problems for the classic solution, while being simpler than the shuntingyard algorithm, is what I call "precedence climbing". (Note, however, that we will climb down the precedence levels.)
Consider the input sequence
a ^ b * c + d + e
The E subroutine of the classic solution will deal with this by three calls to T, and by consuming the 2 "+"s, building a tree
+(+(result of first call, result of second call), result of third call)
We say that this loop directly consumes the two "+" operators.
The precedence climbing algorithm has a similar loop, but it always directly consumes the first binary operator, then it consumes the next binary operator that is of lower precedence, then the next operator that is of lower precedence than that. When it consumes a leftassociative operator, the same loop will also consume the next operator of equal precedence. Let me rewrite the example with operators written at different heights according to their precedence:
+ + * ^ a b c d e
One loop can consume all 4 operators, creating the tree
+(+(*(^(result of first call, result of second call) result of 3rd call), result of 4th call), result of 5th call)
Each operator is assigned a precedence number. To make things more interesting lets add a few more binary operators and use the following precedence tables:


We use the following grammar G3, in which nonterminal Exp is parameterized by a precedence level. The idea is that Exp(p) recognizes expressions which contain no binary operators (other than in parentheses) with precedence less than p
E > Exp(0) Exp(p) > P {B Exp(q)} P > U Exp(q)  "(" E ")"  v B > "+"  ""  "*" "/"  "^"  ""  "&&"  "=" U > ""
The loop implied by the braces, { and }, in the production for Exp(p) presents a problem: when should the loop be exited? This choice is resolved as follows:
In the productions for Exp(p) and P, the recursive use of Exp is parameterized, by a value q. So there is a second choice to resolve: how is q chosen? The value of q is chosen according to the previous operator:
Consider what will happen in parsing the expression, a * b  c * d  e * f = g * h  i * j  k * l. To make things clearer, I'll present this expression 2 dimensionally to show the precedences of the operators:
2 = 3     5 * * * * * * a b c d e f g h i j k l 0 0 0 0
The call to Exp(0) will directly consume exactly the operators indicated by a 0 underneath. The subexpressions: a, b, c*d, e*f, and g*hi*kk*l will be parsed by calls to P and Exp(6), Exp(4), Exp(4) and Exp(3) respectively. The whole parse is illustrated by
In this picture, each call to Exp is indicated by a dashed contour. The number immediately inside the contour indicates the value of the p parameter. Not shown are the calls to P, of which there is one for each variable, in this example.
What about rightassociative operators? Consider an expression
a^b^c
Because of the different way rightassociative operators are treated, Exp(0) will only directly consume the first ^, as the second will be gobbled up by a recursive call to Exp(6).
A recursivedescent parser based on this method looks like this:
Eparser is var t : Tree t := Exp( 0 ) expect( end ) return t
Exp( p ) is var t : Tree t := P while next is a binary operator and prec(binary(next)) >= p const op := binary(next) consume const q := case associativity(op) of Right: prec( op ) Left: 1+prec( op ) const t1 := Exp( q ) t := mkNode( op, t, t1) return t
P is if next is a unary operator const op := unary(next) consume q := prec( op ) const t := Exp( q ) return mkNode( op, t ) else if next = "(" consume const t := Exp( 0 ) expect ")" return t else if next is a v const t := mkLeaf( next ) consume return t else error
I've used precedence climbing in a JavaCC parser for a subset of C++. I've also used it in a parser based on monadic parsing written in Haskell. I'd be happy to mail either grammar to anyone who is interested.
Michael BruceLockhart has implemented a table driven version of the precedence climbing algorithm. Download it here parser.js and parserTest.htm.
Alex de Kruijff has written an implementation of the precedence climbing algorithm as a Java library called Kilmop. You can find it here.
Christian Kaps has created an implementation of the precedence climbing algorithm in PHP. You can find it at https://github.com/mohiva/pyramid.
Terrence Parr has implemented the idea of precedence climbing in version 4 of the ANTLR parser generator. You can read about it in his new book.
As I write this section, it is 2013. I first posted the previous sections over 10 years ago. Lately I've thought a bit more about precedence climbing. I don't want to make the previous section any longer than it already is, so I'm adding this new section. For the interested, it presents an approach to deriving the algorithm by a combination of grammar transformation and program transformation. Keith Clarke did a similar thing long ago [0], but I thought I'd take a crack at it myself. After completing this section, I read Clarke's paper and found that his approach and presentation is almost the same as mine. The main difference is that I generalize the algorithm to handle postfix operators and nonassociative operators. Also, as discussed later, Clarke uses a clever idea to avoid needing the variable I've called "r".
If you are not interested in this, then skip to the next section.
Let's start with a leftrecursive grammar.
S > E0 end E0 > E1  E1 "=" E1 E1 > E2  E1 "+" E2 E2 > E3  E2 "*" E3 E3 > E4  E3 "!" E4 > E5  E5 "^" E4 E5 > P P > "" E2  "(" E0 ")"  v
We have (in order of increasing precedence)
Nonassociative operators aren't found in many languages. That's because we can ban "a=b=c", but it is harder to ban "a=(b=c)". Nevertheless, they are easy to add to the grammar and make the problem just a bit more interesting.
This grammar is not unambiguous, but it almost is. The ambiguity is because "" has lower precedence than "*" and "^". For example, a*b can be parsed two ways: like (a*b) and like (a)*b. I won't worry about the ambiguity for now. I'll just note that the parse we want is the one that is like (a*b); later on, we'll check that our parser really does produce the right tree for a*b. There are other ways of handling unary operators; one advantage of the present approach is that a*b is in the language.
This grammar has both a prefix and postfix operators. But there is a subtle lack of symmetry between them. Note that a^b is in the language of S, whereas b!^a is not. In essence, I'm treating postfix operators as binary operators that have no right operand. We'll come back to this issue at the very end of this section. Meanwhile, we'll leave the grammar as is.
Note that we can use the same symbol for a prefix or binary operators. (E.g., making '' a binary operator as well as prefix is perfectly fine.) However, I'll assume that the sets of postfix and binary operators don't overlap. The reason is that it is easy to distinguish a unary from a binary operator based on the left context: unary operators follow the start of the expression, left parentheses, binary operators, and unary operators. It is not so easy to tell a binary operator from a postfix operator. Also if we allow operators to be unary, binary and postfix, there is more ambiguity. For example ab is unambiguous if  is unary and binary, but ambiguous, if  is also postfix.
There is a transformation rule that A > a  A b can be written as A > a {b} where {b} means 0 or more b. We can apply rule this to nonterminals E1, E2, and E3.
Another rule says we can rewrite A > s  s b as A > s [b], where [b] means 0 or 1 b. We apply this to nonterminals E0 and E4.
The result is
S > E0 end E0 > E1 [ "=" E1 ] E1 > E2 { "+" E2 } E2 > E3 {"*" E3 } E3 > E4 { "!" } E4 > E5 [ "^" E4 ] E5 > P P > "" E2  "(" E0 ")"  v
At this point, if we convert to recursive descent code, we get the "classic" algorithm.
Next we substitute righthand sides for lefthand sides. We start by rewriting E4's rule to E4 > P ["^" E4]. Then E3 can be rewritten as E3 > P ["^" E4] { "!" } and we work backwards like this to E0. The result is
S > E0 end E0 > P [ "^" E4 ] { "!" } {"*" E3 } { "+" E2 } [ "=" E1 ] E1 > P [ "^" E4 ] { "!" } {"*" E3 } { "+" E2 } E2 > P [ "^" E4 ] { "!" } {"*" E3 } E3 > P [ "^" E4 ] { "!" } E4 > P [ "^" E4 ] P > "" E2  "(" E0 ")"  v
(E5 is no longer reachable, so I trashed it.)
It's a hard to believe we are making progress; bear with me.
Now we want to take advantage of the similarity of the rules for E0 through E4. To do this we need a bit more notation. First we'll use a parameterized nonterminal E: so E(n) below will match the same strings as En above. Second, I'll use the notation ?(B), where B is a boolean expression, as a "test". The meaning of a test is that a particular alternative of the grammar can only be taken when B is true. For example X [ ?(B) Y] means the same as X [Y] when B is true, but means the same as X when B is false. Here is the compacted grammar.
S > E(0) end E(p) > P [ ?(p≤4) "^" E(4) ] { ?(p≤3) "!" } { ?(p≤2) "*" E(3) } { ?(p≤1) "+" E(2) } [ ?(p≤0) "=" E(1) ] P > "" E(2)  "(" E(0) ")"  v
Now we are getting somewhere.
The fourth transformation is easier to explain as a code transformation than as a grammar transformation, so, to get ready, we convert our grammar to a recognizer using the usual recursive descent magic. I won't add any tree building commands yet, as they clutter things up.
S is E(0) expect( end ) E(p) is P if p≤4 and next="^" then ( consume ; E(4) ) while p≤3 and next="!" do consume while p≤2 and next="*" do ( consume ; E(3) ) while p≤1 and next="+" do ( consume ; E(2) ) if p≤0 and next="=" then ( consume ; E(1) ) P is if next="" then ( consume ; E(2) ) else if next = "(" then ( consume ; E(0) ; expect( ")" ) ) else if next is a v then consume else error
This is a deterministic algorithm for an ambiguous grammar: While converting from a grammar to a set of subroutines, I've made a decision about what happens in the ambiguous cases. So, even though there are no tree building commands, at this point we should stop and consider strings such as a*b and a+b, and see whether they are parsed "correctly". I.e., we should ask: if we did have tree building commands, would the right tree be built? Here is what happens for a*b . S calls E(0), E(0) calls P, and P consumes the "" and then calls E(2). After consuming "a", the call to E(2) sees the "*". The crucial decision is to consume the "*" and the "b" in the loop, rather than returning from E(2) as soon as "a" is consumed. The right tree is built  or would be, if we were building trees. By contrast, in parsing a+b, E(2) consumes the "a" and then must return; so it is the call to E(0) that consumes the "+". The tree we would get is the same as that for (a)+b, just as desired.
The next transformation is the trickiest one. We want to combine the sequence of if commands and while loops in procedure E into one single while loop. If I have two identical loops sequentially, say
while A do X while A do X ,
we can rewrite them as one loop
while A do X
What if the loops are different?. Suppose we have two successive while commands
while A do X while B do Y ,
and, furthermore, know that not A is a loop invariant of the second loop (i.e. that not A and B implies that executing Y leaves A false), then we can rewrite to
while A or B do if A then X else Y
Similarly, if we have an if command followed by a while command
if A then X while B do Y ,
and, furthermore, know that not A is an invariant of the loop and that after executing X, A will be false, then we can rewrite to
while A or B do if A then X else Y
What if there is no special relationship between the conditions and the commands? Then we can create one. We can use a counter to keep track of how many of the original loops and ifs our one loop is still emulating: For example
if A do X while B do Y while C do Z
can always be rewritten as
var r := 2 while 2≤r and A or 1≤r and B or 0≤r and C do if 2≤r and A then (X ; r := 1) else if 1≤r and B then (Y ; r := 1) else (Z ; r := 0)
where r is a fresh variable. The r variable acts as a ratchet; it prevents backsliding. Once X is executed, we set r to ensure X is never executed again. Once Y is executed, we set r to ensure that X is never executed in the future. Once Z is executed, we set r to ensure that neither X nor Y is executed in the future.
Applying this loop fusing idea to our latest E procedure we get
E(p) is P var r := 4 while p≤4≤r and next="^" or p≤3≤r and next="!" or p≤2≤r and next="*" or p≤1≤r and next="+" or p≤0≤r and next="=" do if p≤4≤r and next="^" then (consume ; E(4) ; r := 3) else if p≤3≤r and next="!" then (consume ; r := 3) else if p≤2≤r and next="*" then (consume ; E(3) ; r := 2) else if p≤1≤r and next="+" then (consume ; E(2) ; r := 1) else /* p≤0≤r and next="=" */ (consume ; E(1) ; r := 1)
Take a look at the latest E procedure. If we could eliminate the differences between the branches of the nested ifs, we wouldn't need any branching within the loop body. If we could eliminate the differences between the disjuncts of the loop guard, we could greatly simplify the loop guard. The differences come down to whether an operator is binary or postfix and three numbers:
For example, for + these numbers are respectively 1 (from p≤1≤r and next="+"), 2 (from E(2)), and 1 (from r := 1).
We make the following three tables.
b  =  +  *  !  ^ 

prec(b)  0  1  2  3  4 
rightPrec(b)  1  2  3  NA  4 
nextPrec(b)  1  1  2  3  3 
Now we rewrite the latest E procedure to use the tables and, after simplifying, we get
E(p) is P var r := 4 while next is a binary or a postfix operator and p≤prec(next)≤r do const b := next consume if b is binary then E( rightPrec(b) ) r := nextPrec(b)
This transformation is the one that really makes the algorithm both compact and efficient. The number of operations executed and the size of the code (excluding the table) are now both independent of the number of precedence levels.
We don't really need three tables. We can define rightPrec(b)=1+prec(b) when b is leftassociative or nonassociative, and rightPrec(b)=prec(b) when b is rightassociative. We can define nextPrec(b)=prec(b) when b is leftassociative or postfix, and nextPrec(b)=prec(b)1 when b is rightassociative or nonassociative. (If we have a postfix operator $ and want to prevent, say, a$$", we make nextPrec('$')=prec('$')1.
The really alert reader may notice that we didn't have the r variable or the nextPrec function in the previous section. We need them here because we have nonassociative operators and postfix operators. It also simplifies the argument that the transformation is correct. I'll leave it as an exercise to eliminate r when it's not needed. (Or you can read Clarke's paper [0].)
In the case of postfix operators, the use of r and nextPrec ensures that, for example, a!^b is an error, just as it is in the original grammar. If we want to allow such strings, we can set nextPrec('!') to 4. In the first version of procedure E, this amounts to a backwards jump like this:
E(p) is P L: if p≤4 and next="^" then ( consume ; E(4) ) if p≤3 and next="!" then ( consume ; goto L) while p≤2 and next="*" do ( consume ; E(3) ) while p≤1 and next="+" do ( consume ; E(2) ) if p≤0 and next="=" then ( consume ; E(1) )
In general, the nextPrec for any postfix operator can be the precedence of the highest precedence operator. This allows a postfix operator to be followed by any binary or postfix operator, just as a prefix operator can follow any binary or prefix operator.
Finally, we add the tree building operations. Strictly speaking they should have been added as soon as we went to code, so that we could see that the right tree is built for ambiguous inputs and that subsequent transformations have no effect on the tree that is finally built.
S is const t := E(0) ; expect( end ) ; output(t) E(p) is var t := P var r := 4 while next is a binary or a postfix operator and p≤prec(next)≤r do const b := next consume if b is binary then const t1 := E( rightPrec(b) ) t := mknode(binary(b), t, t1) else t := mknode(postfix(b), t) r := nextPrec(b) return t P is if next="" then ( consume ; const t:= E(2) ; return mknode(prefix('', t)) ) else if next = "(" then ( consume ; const t := E(0) ; expect(")") ; return t ) else if next is a v then ( const t := mkleaf(next) ; consume ; return t ) else error
Recursive descent seems to have been first described by Peter Lucas, who, with a team from IBM's Vienna laboratory, used it in their ALGOL 60 compiler. In his report [2], he writes
The translator will be designed in such a way, that each meaning of a metalinguistic variable corresponds to a subroutine which carries out the transformation of the string of symbols.
Another early use of recursive descent parsing was in an ALGOL 60 Compiler for the Elliott Brothers' 803 and 503 computers; it was designed by a team of three programmers led by C. A. R. Hoare. From Hoare's report on that compiler:
The main work is done by a set of procedures, each of which is capable of processing one of the syntactic units defined in the ALGOL 60 report. Where one syntactic unit is defined as consisting of other units, the procedure will be capable of activating other procedures, and where necessary, itself. For example, the procedure "compile arithmetic expression" must be capable of compiling the bracketed constituents of an arithmetic expression, which are themselves arithmetic expressions; this is achieved by a recursive entry to the very procedure "compile arithmetic expression" which is currently engaged on translating the whole expression. [3]
You can use this compiler yourself; see http://elliott803.sourceforge.net/docs/algol.html. You can read the (disassembled) compiler at http://www.billp.org/ccs/ElliottAlgol/.
I'm not sure who invented what I am calling the classic algorithm. (Anyone know?) It was made popular, I think, by Niklaus Wirth who used it in various compilers, notably for Pascal. I learned it from one of Wirth's books.
The ShuntingYard Algorithm was apparently invented by Edsger Dijkstra around 1960 in connection with one of the first ALGOL 60 compilers. It is described in a Mathematisch Centrum report (starting around page 21) [1]. I say "aparently", as essentially the same algorithm is described by Friedrich. Bauer and Klaus Samelson in 1960 [5]. I think I first saw a version of it described in an ad for the TI SR52 and SR56 calculators, two of the earliest pocket calculators to handle precedence. (Prior to 1976, pocket calculators either used RPN or treated 2+3*4 as (2+3)*2.)
I first saw what I've called the precedence climbing method described by Keith Clarke in a posting to comp.compilers in 1992. Keith gives a proof of its correctness, relative to the classic algorithm, by means of program transformation in a 1985 report "The topdown parsing of expressions" [0]. The algorithm appears to have been first invented by Martin Richards for use in the CPL and BCPL compilers. It can be found in the section 6.6 of BCPL  the language and its compiler [4] and can still be found in recent distributions of BCPL's compiler. Neither Clarke nor Richards had a special name for the algorithm, so my one contribution (besides extending it to postfix and nonassociative operators) is suggesting the name "precedence climbing".
It turns out that precedence climbing is a special case of a more flexible technique called Pratt parsing. Pratt parsing was first described in a paper by Vaughn Pratt [6]. This connection was brought to my attention by Andy Chu who wrote about it in a blog post [7]. I've explored this connection in my own web article From Precedence Climbing to Pratt Parsing [8].
[0] Keith Clarke, "The topdown parsing of expressions", Research Report 383, Dept of Computer Science, Queen Mary College. Archived at http://antlr.org/papers/Clarkeexprparsing1986.pdf.
[1] Edsger W. Dijkstra, "Algol 60 translation : An Algol 60 translator for the x1 and Making a translator for Algol 60", Research Report 35, Mathematisch Centrum, Amsterdam, 1961. Reprint archived at http://www.cs.utexas.edu/users/EWD/MCReps/MR35.PDF.
[2] Peter Lucas, "The Structure of FormulaTranslators", ALGOL Bulletin, Issue Sup 16, Sep. 1961. Archived at http://archive.computerhistory.org/resources/text/algol/algol_bulletin/AS16/AS16.HTM.
[3] C. A. R. Hoare, "Report on the Elliott ALGOL translator", The Computer Journal, Vol. 5, No. 2, pp. 127129, 1962. Archived at http://comjnl.oxfordjournals.org/content/5/2/127.short.
[4] Martin Richards and Collin WhitbyStevens, BCPL  the language and its compiler, 1st ed., Cambridge University Press, 1979.
[5] F. L. Bauer and K. Samelson, "Sequential formula translation", Communications of the ACM, vol 3, #2, February, 1960.
[6] Vaughn R. Pratt, "Top down operator precedence", Proceedings of the 1st symposium on principles of programming languages (POPL), 1973, http://dl.acm.org/citation.cfm?id=512931
[7] Andy Chu, "Pratt Parsing and Precedence Climbing Are the Same Algorithm", Nov 2016. http://www.oilshell.org/blog/2016/11/01.html
[8] Theodore S. Norvell, "From Precedence Climbing to Pratt Parsing", 2016, www.engr.mun.ca/~theo/Misc/pratt_parsing.htm
Thanks to Colas Schretter for pointing out an error in the precedence climbing algorithm and suggesting a correction.
I am grateful to Keith Clarke and Martin Richards for helping me trace the origins of what I've called precedence climbing and to Andy Chu for pointing out the connection to Pratt parsing.
Thanks to everyone who took the trouble to email me to tell me that they found this article useful, to point me to an implementation, or to tell me that I've made an error.