How does preprocessing work in this snippet?

Question

#include<stdio.h>
#define A -B
#define B -C
#define C 5

int main() {
  printf("The value of A is %dn", A); 
  return 0;
} 

I came across the above code. I thought that after preprocessing, it gets transformed to

// code from stdio.h

int main() {
  printf("The value of A is %dn", --5); 
  return 0;
}

which should result in a compilation error. But, the code compiles fine and produces output 5.

How does the code get preprocessed in this case so that it does not result into a compiler error?

PS: I am using gcc version 8.2.0 on Linux x86-64.

Answer 1

The preprocessor is defined as operating on a stream of tokens, not text. You have to read through all of sections 5.1.1, 6.4, and 6.10 of the C standard to fully understand how this works, but the critical bits are in 5.1.1.1 "Phases of translation": in phase 3, the source file is "decomposed into preprocessing tokens"; phases 4, 5, and 6 operate on those tokens; and in phase 7 "each preprocessing token is converted into a token". That indefinite article is critical: each preprocessing token becomes exactly one token.

What this means is, if you start with this source file

#define A -B
#define B -C
#define C 5
A

then, after translation phase 4 (macro expansion, among other things), what you have is a sequence of three preprocessing tokens,

<punctuator: -> <punctuator: -> <pp-number: 5>

and at the beginning of translation phase 7 that becomes

TK_MINUS TK_MINUS TK_INTEGER:5

which is then parsed as the expression -(-(5)) rather than as --(5). The standard offers no latitude in this: a C compiler that parses your example as --(5) is defective.

When you ask a compiler to dump out preprocessed source as text, the form of that text is not specified by the standard; typically, what you get has whitespace inserted as necessary so that a human will understand it the same way translation phase 7 would have.