Thread: Why is the answer 5 to this code and not 4?

I am still a novice. Can someone explain this to me? Why is the answer 5 instead of 4?

insert

Code:

#include<stdio.h>
#define TWICE(x) 2*x
main(){
printf("%d\n", TWICE(3-1));
}

Actually, that still evaluates to 4. Did you mean to say that it expands to 2*3-1, which is equal to 5?

Simple:

Code:

#define TWICE(x) 2*x
main(){
printf("%d\n", TWICE(3-1));
}

After preprocessor processing, leaving off "#include<stdio.h>"

Code:

main(){
printf("%d\n", 2*3-1);
}

2 times 3 minus 1 is 5!

Your code should look something like:

Code:

#include<stdio.h>

#define TWICE(x) 2 * (x)

int main(void)
{
   printf("%d\n", TWICE(3-1));
}

Please note the added parentheses around the second 'x' in the #define! The 'x' in the #define is just a textual placeholder.