Q33
What will be the output of the program? #include #define int char void main() int i = 65; printf("sizeof(i)=%d", sizeof(i));
A.
sizeof(i)=2
B.
sizeof(i)=1
AnswerC.
Compiler Error
D.
None of These
Answer: Option B
Solution
Answer: Option B
Solution:
Since the #define replaces the string int by the macro char.
So, here i is a variable of type char and not int.