What will be the output of the program?
#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.

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