The expected output of the following C program is to print the elements in the array. But when actually run, it doesn't do so.
#include<stdio.h>
#define TOTAL_ELEMENTS (sizeof(array) /
sizeof(array[0]))
int
array[] = {23,34,12,17,204,99,16};
int
main()
{
int
d;
for(d=-1;d
<= (TOTAL_ELEMENTS-2);d++)
printf("%d\n",array[d+1]);
return
0;
}
Find out what's going wrong.
I thought the following program was a perfect C program. But on compiling, I found a silly mistake. Can you find it out (without compiling the program :-) ?
#include<stdio.h>
void
OS_Solaris_print()
{
printf("Solaris - Sun Microsystems\n");
}
void
OS_Windows_print()
{
printf("Windows - Microsoft\n");
}
void
OS_HP-UX_print()
{
printf("HP-UX - Hewlett Packard\n");
}
int
main()
{
int
num;
printf("Enter the number (1-3):\n");
scanf("%d",&num);
switch(num)
{
case
1:
OS_Solaris_print();
break;
case
2:
OS_Windows_print();
break;
case
3:
OS_HP-UX_print();
break;
default:
printf("Hmm! only 1-3 :-)\n");
break;
}
return
0;
}
What's the expected output for the following program and why? enum {false,true};
int {
{
}while(false);
}
The following program doesn't "seem" to print "hello-out". (Try executing it)
What could be the reason?
|
#include <stdio.h>
#define f(a,b) a##b
#define g(a) #a
#define h(a) g(a)
int
main()
{
printf("%s\n",h(f(1,2)));
printf("%s\n",g(f(1,2)));
return
0;
}
Just by looking at the program one "might" expect the output to be, the same for both the printf statements. But on running the program you get it as:
bash$ ./a.out
12
f(1,2)
bash$
Why is it so?
