Why #define INVALID_VALUE -999; give syntax error when using?

Question

I am trying to compare with specific constants in C, and I simplified my program as follows:

#include "stdio.h"
#include "stdlib.h"
#define INVALID_VALUE -999;

int main(void)
{
    int test=0;
    if(test==INVALID_VALUE) //The error line..
        return INVALID_VALUE;
    return 0;
}

And when I use gcc to compile, it throws an error " ". error: expected ‘)’ before ‘;’ token

Is there a reason this cannot be done?

Answer 1

Remove the semicolon from the INVALID_VALUE definition.

Macros are replaced lexically (symbolically) without understanding the syntax around them. Your macro INVALID_VALUE is set -999;, so your if line extends the macro to:

if (test==-999;)

which is invalid C syntax

Answer 2

; #define INVALID_VALUE -999;. . .

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Answer 3

- C C

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Answer 4

Half Column at the End

#define INVALID_VALUE -999;

Classical.

Answer 5

You do not need a semicolon after defining something. #define is actually a macro, and it will do an inline extension when compiling.

In this way,

#define IDENTIFIER 10;
int j = IDENTIFIER;

will expand as:

int j = 10;;

Answer 6

Change macro

from

#define INVALID_VALUE -999;

to

#define INVALID_VALUE -999

Bye