#define x 2|0 in C

Question

The below given code is written to satisfy the condition (x == x+2) to be returning true in C.

#include<stdio.h>
#define x 2|0

int main()  
{
    printf("%d",x==x+2); 
    return 0;
}

In the above code why the printf() is printing 2 ( if I write x+3 I get 3 and so on ).

Can someone explain how the given macro is working.

What is the use of | operator in C and what does the macro

#define x 2|0

mean? I read about macros in other questions but no question explained similar kind of example.

Answer 1

TL;DR; Read about operator precedence.

+ binds higher than == which binds higher than |.

After preprocessing, your printf() statement looks like

 printf("%d",2|0==2|0+2);

which is the same as

 printf("%d",2|(0==2)|(0+2));

which is

printf("%d",2|0|2);

---

Word of advice: Do not write this type of code in real scenario. With minimal level of compiler warning enabled, your code produces

source_file.c: In function ‘main’:
source_file.c:4:12: warning: suggest parentheses around comparison in operand of ‘|’ [-Wparentheses]
 #define x 2|0
            ^
source_file.c:8:21: note: in expansion of macro ‘x’
     printf("%d\n\n",x==x+2); 
                     ^
source_file.c:4:12: warning: suggest parentheses around arithmetic in operand of ‘|’ [-Wparentheses]
 #define x 2|0
            ^
source_file.c:8:24: note: in expansion of macro ‘x’
     printf("%d\n\n",x==x+2);

So, the moment you change the MACRO definition to something sane, like

#define x (2|0)

the result will also change, as the explicit precedence then will be guaranteed by the parenthesis.