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 , used in the function call sum(int, int) is not a comma operator?

NOTE: I didn't want to actually use the comma operator in the function call. I just wanted to know how the compiler knows that it is not a comma operator.

Answer 1

Look at the grammar for the C language. It's listed, in full, in Appendix A of the standard. The way it works is that you can step through each token in a C program and match them up with the next item in the grammar. At each step you have only a limited number of options, so the interpretation of any given character will depend on the context in which it appears. Inside each rule in the grammar, each line gives a valid alternative for the program to match.

Specifically, if you look for parameter-list, you will see that it contains an explicit comma. Therefore, whenever the compiler's C parser is in "parameter-list" mode, commas that it finds will be understood as parameter separators, not as comma operators. The same is true for brackets (that can also occur in expressions).

This works because the parameter-list rule is careful to use assignment-expression rules, rather than just the plain expression rule. An expression can contain commas, whereas an assignment-expression cannot. If this were not the case the grammar would be ambiguous, and the compiler would not know what to do when it encountered a comma inside a parameter list.

However, an opening bracket, for example, that is not part of a function definition/call, or an if, while, or for statement, will be interpreted as part of an expression (because there's no other option, but only if the start of an expression is a valid choice at that point), and then, inside the brackets, the expression syntax rules will apply, and that allows comma operators.