PreprocessorWhat will be the output of the following C code?#include <stdio.h>#define int charvoid main() { int i = 65; printf("%d", sizeof(i));}Options42Compilation Error1

The output of the given C code will be 1.

Here's the step by step explanation:

1. The #define directive in C is a pre-processor directive which is used to define symbolic names or macros. In the given code, #define int char is replacing all occurrences of int with char before the code is compiled.

2. So, the line int i = 65; is actually interpreted as char i = 65; by the compiler.

3. The sizeof operator in C is used to calculate the size of variables, constants, arrays, structures etc. The size of char is always 1 byte in C.

4. Therefore, printf("%d", sizeof(i)); will print 1, because the size of i (which is a char due to the #define directive) is 1 byte.