How does the compiler know that the comma in a function call is not a comma operator?

Question

Consider the function call (calling int sum(int, int))

printf(\"%d\", sum(a,b));

How does the compiler decide that the

Answer 1

The reason is the C Grammar. While everyone else seems to like to cite the example, the real deal is the phrase structure grammar for function calls in the Standard (C99). Yes, a function call consists of the () operator applied to a postfix expression (like for example an identifier):

 6.5.2 postfix-expression:
       ...
       postfix-expression ( argument-expression-list_opt )

together with

argument-expression-list:
       assignment-expression
       argument-expression-list , assignment-expression    <-- arglist comma

expression:
       assignment-expression
       expression , assignment-expression                  <-- comma operator

The comma operator can only occur in an expression, i.e. further down the in the grammar. So the compiler treats a comma in a function argument list as the one separating assignment-expressions, not as one separating expressions.