C symbolic constant wrong value

Asked Jun 12 '21 at 20:11

Active Jun 12 '21 at 20:12

Viewed 32 times

I am a newbie in C. I am wondering why the following CONST2 is being printed as 13 instead of 18:

#include <stdio.h>


#define CONST 2 + 3 + 4
#define CONST2 CONST * 2


int main()
{
  printf("%d %d\n", CONST, CONST2);
  return 0;
}

which using gcc compiler prints:

/tmp $ gcc a.c 
/tmp $ ./a.out 
9 13

Any help is appreciated.

edited Jun 12 '21 at 20:12

asked Jun 12 '21 at 20:11

Student

Hint: `CONST2` expands to `2 + 3 + 4 * 2`. https://en.cppreference.com/w/c/language/operator_precedence You probably want `#define CONST (2 + 3 + 4)`. – Nate Eldredge Jun 12 '21 at 20:12
1

because 2+3+4*2 is 13 – Pablo Jun 12 '21 at 20:12
Means #defined symbols are replaced exactly with the string of their values, rather than with the actual value? – Student Jun 12 '21 at 20:14
@Student: Correct, more or less. – Nate Eldredge Jun 12 '21 at 20:15
2

@Student yes, macros are replaces exactly as they were written. – Pablo Jun 12 '21 at 20:15
1

To be precise, the string *is* the actual value of the macro, at this point in compilation. – IS4 Jun 12 '21 at 20:34
1

macros are texts replacement, it happens before compilation so obviously it knows nothing about expressions. duplicate: [The need for parentheses in macros in C](https://stackoverflow.com/q/10820340/995714) – phuclv Jun 13 '21 at 02:25

0 Answers0