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; } ?->(Show Answer!)

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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; } ?->(Show Answer!)

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1. 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; }

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By: guest on 01 Jun 2017 06.01 pm

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\n", i); It prints the value of variable i.(5) Hence the output is '5'.

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