Why my code giving a wrong output?

Question

I am running the code below:

#include<stdio.h>
#define x 5+2

int main() {

    int p;
    p = x*x*x;
    printf("%d",p);

}

I was expecting 343 as the output, but the answer is 27. Why so? I am unable to figure it out why it's happening. Please help.

Answer 1

This is why it's 27 and not 343: . before -! So look at this:

#include<stdio.h>
#define x 5+2

int main() {

    int p;
    p = x*x*x;  //5 + 2*5 + 2*5 + 2 -> 5 + (2*5) + (2*5) + 2 -> 5 + 10 + 10 + 2 = 27

    printf("%d",p);

    return 0;

}

You would have to change it to this:

p = (x)*(x)*(x);

Or define it with this line:

#define x (5+2)

So that the result is 343

As a reference take a look here to the C Operator precedene: http://en.cppreference.com/w/c/language/operator_precedence

Answer 2

p = 5+2*5+2*5+2

ans: 27

use:

#define x (5+2)

Answer 3

x * x * x with #define x 5 + 2 will expand to

p = 5 + 2 * 5 + 2 * 5 + 2

and hence you will have 5 + 10 + 10 + 2 = 27, you have to add parentheses

#define x (5 + 2)

and then it will expand to

p = (5 + 2) * (5 + 2) * (5 + 2) // -> 7 * 7 * 7 -> 343

#define does not define variables, it defines macros.

You can read about the c preprocessor for more information.

Answer 4

After macro expansion

p=x*x*x;  

will be

p = 5+2*5+2*5+2;