Predict the output of following C programs.
Question 1:
C
#include<stdio.h>
int main()
{
enum channel {star, sony, zee};
enum symbol {hash, star};
int i = 0;
for (i = star; i <= zee; i++)
{
printf ( "%d " , i);
}
return 0;
}
|
Output:
compiler error: redeclaration of enumerator 'star'
In the above program, enumeration constant ‘star’ appears two times in main() which causes the error. An enumeration constant must be unique within the scope in which it is defined. The following program works fine and prints 0 1 2 as the enumeration constants automatically get the values starting from 0.
C
#include<stdio.h>
int main()
{
enum channel {star, sony, zee};
int i = 0;
for (i = star; i <= zee; i++)
{
printf ( "%d " , i);
}
return 0;
}
|
Output:
0 1 2
Question 2:
C
#include<stdio.h>
int main()
{
int i, j;
int p = 0, q = 2;
for (i = 0, j = 0; i < p, j < q; i++, j++)
{
printf ( "GeeksforGeeks\n" );
}
return 0;
}
|
Output:
GeeksforGeeks
GeeksforGeeks
Following is the main expression to consider in the above program.
When two expressions are separated by comma operator, the first expression (i < p) is executed first. Result of the first expression is ignored. Then the second expression (j < q) is executed and the result of this second expression is the final result of the complete expression (i < p, j < q). The value of expression ‘j < q’ is true for two iterations, so we get “GeeksforGeeks” two times on the screen. See this for more details.
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