Comments by Aswin Waiba (Top 14 by date)

Try.

int a()
{
printf("\nHello World\n");
return 1; //or 0 your call
}

Your program should work. And remember when working with function pointers:
typedef int (*foo)();
| | |
return type| parameters
typename
so for your original function void a(); it should have been
typedef void (*foo)();
foo fnptr=&a; //& not required

but since you are checking for a value inside if statement while calling the function. You probably wanted a function that returned some value.
Since your original function is void the if condition will always be false and the else part will be executed.

A better way to go about this would be:
int a(int param)
{
printf("Hello World\n");
return param;
}

int main()
{
typedef int (*Fun_Ptr)(int);
Fun_Ptr funPtr=&a;

int aNumber;
printf("\nEnter a number :");
scanf("%d",&aNumber);
if(50>funPtr(aNumber))
printf("\nOMG!!This is working");
else
printf("\nThis was supposed to work");
}

Its really simple eg and we are doing something useless such as pass a parameter to the function and returning it back unchanged. You could perform some sort of opertion on it like modulo or some thing. Anyway the idea is to understand the syntax for declaring a function pointer properly and using it properly.
Though the compiler will allow you to run your code its not safe and this kind of behaviour in the code should be avoided to all best intents.

I hope this is Clear.