Why the output for “a” is -80?

Question

#include<stdio.h>
#include<conio.h>
#define ABC 20
#define XYZ 10
#define XXX ABC - XYZ

void main()
{
    int     a;

    a = XXX * 10;

    printf("\n %d \n", a);

    getch();
}

I thought the output should be 100 but when I saw the result I found output as -80. When I put brackets as #define XXX (ABC-XYZ) then I get output as 100 but without brackets I get output as -80.

Answer 1

The preprocessor is a dumb beast. It works on a purely textual level, without regard for what it's doing to the code semantically. Let's see what it does in this case:

a = XXX * 10;

becomes

a = ABC - XYZ * 10;

which becomes

a = 20 - 10 * 10;

which is, obviously, -80.

What was probably intended was

#define XXX (ABC - XYZ)

This is one reason why macros are considered evil.

Answer 2

a = XXX * 10;

will change to the following before compilation:

a = 20 - 10 * 10;

which evaluates to -80

Answer 3

It is calculating like this ,

20-10*10

Use braces in the macro XXX.

#define XXX (ABC-XYZ)

Answer 4

Here XXX is replaced by ABC-XYZ So it is look like the follow

20 - 10 * 10

So the output is -80