Predict the output of following C programs.
Question 1
#include<stdio.h> #define R 10 #define C 20 int main()
{ int (*p)[R][C];
printf ( "%d" , sizeof (*p));
getchar ();
return 0;
} |
Output: 10*20*sizeof(int) which is “800” for compilers with integer size as 4 bytes.
The pointer p is de-referenced, hence it yields type of the object. In the present case, it is an array of array of integers. So, it prints R*C*sizeof(int).
Thanks to Venki for suggesting this solution.
Question 2
#include<stdio.h> #define f(g,g2) g##g2 int main()
{ int var12 = 100;
printf ( "%d" , f(var,12));
getchar ();
return 0;
} |
Output: 100
The operator ## is called “Token-Pasting” or “Merge” Operator. It merges two tokens into one token. So, after preprocessing, the main function becomes as follows, and prints 100.
int main()
{ int var12 = 100;
printf ( "%d" , var12);
getchar ();
return 0;
} |
Question 3
#include<stdio.h> int main()
{ unsigned int x = -1;
int y = ~0;
if (x == y)
printf ( "same" );
else
printf ( "not same" );
printf ( "\n x is %u, y is %u" , x, y);
getchar ();
return 0;
} |
Output: “same x is MAXUINT, y is MAXUINT” Where MAXUINT is the maximum possible value for an unsigned integer.
-1 and ~0 essentially have same bit pattern, hence x and y must be same. In the comparison, y is promoted to unsigned and compared against x. The result is “same”. However, when interpreted as signed and unsigned their numerical values will differ. x is MAXUNIT and y is -1. Since we have %u for y also, the output will be MAXUNIT and MAXUNIT.
Thanks to Venki for explanation.
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