GeeksforGeeks » C/C++ Programming Questions
A hedge fund interview question: what's the output result? And explain why.
(5 posts)-
what's the output result? And explain why.
#include <stdio.h>
int main()
{
char *p;
char buf[10] = {1, 2, 3, 4, 5, 6, 9, 8};
p = (buf+1)[5];
printf("%d \n", p);
} -
I got compilation error in below statement
p = (buf+1)[5];
The error message
"invalid conversion from
char' tochar*' " -
I think the code can be corrected by putting a * before p. Like below code works and prints 9.
#include <stdio.h>
int main()
{
char *p;
char buf[10] = {1, 2, 3, 4, 5, 6, 9, 8};
*p = (buf+1)[5];
printf("%d \n", *p);
getchar();
return 0;
} -
As I was emphasizing in previous posts, the code will not compiler due to type mismatch.
The expression (buf+1) evaluates to pointer to char type and (buf+1)[5] evaluates to 5th 'char'acter of array pointed by the pointer (buf+1). Hence resulting type of the expression "(buf+1)[5]" is char, which we assigned to char pointer. This conversion is unusual in C/C++ programming. Results in compilation error. If you use explicit cast from char to char * to get rid of error, the code fails at runtime.
-
I think the correct way is
int main(){
char *p;
p=(char*)malloc(sizeof(char));
char buf[10] = {1, 2, 3, 4, 5, 6, 9, 8};
*p = (buf+1)[5];
printf("%d \n", *p);
getch();}
o/p-9
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