What will be the output of the program?
#include<stdio.h>
#define int char
void main()
{
int i = 65;
printf("sizeof(i)=%d", sizeof(i));
}
#include<stdio.h>
#define int char
void main()
{
int i = 65;
printf("sizeof(i)=%d", sizeof(i));
}A. sizeof(i)=2
B. sizeof(i)=1
C. Compiler Error
D. None of These
Answer: Option B
Solution (By Examveda Team)
Since the #define replaces the string int by the macro char.
So, here i is a variable of type char and not int.
This Question Belongs to C Program >> C Preprocessor
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