Question

What is accomplished by the call to sprintf in the codefragment below?

Answer 1

Answer:

E. None of the above

Explanation:

char ans[20]

int num = 40;

sprintf(ans, "%d to %d", num, num +10);

This code fragment produce an error,because in 1st line of code char ans[20]  ,there is no semicolon after the line.As ans is undeclared ,the ide will produce an error as we use it in

                   sprintf(ans, "%d to %d", num, num +10);

If we put semicolon after the char ans[20] ,then it will not give error but also don't produce an output because for sprintf we should initialize the num and ans as of pointer type.

It is a condition for sprintf and num should be of string type.

C program

#include <stdio.h>

int main()  

{  

char ans[20];  

int num = 40;  

sprintf(ans, "%d to %d", num, num +10);  

return 0;  

}

Output  

No output  

(You can check this code on any ide for confirmation.)