Question
#include<stdio.h>
int fun(int i)
{
i++;
return i;
}
int main()
{
int fun(int);
int i=3;
fun(i=fun(fun(i)));
printf("%d\n", i);
return 0;
}
What will be the output of the program?
#include<stdio.h>
int fun(int i)
{
i++;
return i;
}
int main()
{
int fun(int);
int i=3;
fun(i=fun(fun(i)));
printf("%d\n", i);
return 0;
}
Answer: Option A
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Step 1: int fun(int); This is prototype of function fun(). It tells the compiler that the function
fun() accept one integer parameter and returns an integer value.
Step 2: int i=3; The variable i is declared as an integer type and initialized to value 3.
Step 3: fun(i=fun(fun(i)));. The function fun(i) increements the value of i by 1(one) and return it.
Lets go step by step,
=> fun(i) becomes fun(3) is called and it returns 4.
=> i = fun(fun(i)) becomes i = fun(4) is called and it returns 5 and stored in variable i.(i=5)
=> fun(i=fun(fun(i))); becomes fun(5); is called and it return 6 and nowhere the return value is stored.
Step 4: printf("%d`setminus`n", i); It prints the value of variable i.(5)
Hence the output is '5'.
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