how exactly stack unwinding works between caller function and called function

Question

I have two small functions called from main. Below is the program

int fun1(int x)
{
   int s;
   s=8;
}

int fun2(int x)
{
   int s;
   s=x*4;
}
int main(int argc, char *argv[])
{
   int n=2;
   cout << fun1(n)<<endl;
   cout<<fun2(n)<<endl;
   return 0;
}

1.fun1() function called with arguement x but no operation done using x variable. So in the output "cout<< fun1()-- I am getting output 2 "present value of n".

2.fun2() function called with arguement x and used it in operation "s=x*4" and in the output "cout<< fun2()"-- I am getting output 8--"2*4".

I am not passing any reference or pointer value to the function still getting two different outputs. Would like to know how the stack unwinding is different in the two different function cases.

Is the absence of return values from your functions a typo or is this the genuine code?? — mathematician1975, Dec 21 '13 at 14:25
@mathematician1975 yes. this is the genuine code. I compiled it in windriver rtos with g++ version 2.6.25.5. — Santosh Sahu, Dec 21 '13 at 14:26
Since your code has undefined behaviour, it isn't "doing" anything that we can discuss. — Kerrek SB, Dec 21 '13 at 14:35

score 1 · Answer 1 · edited May 23 '17 at 11:50

1

Your functions do not return anything , despite having a return type. This is undefined behaviour. This question/answer provides more details Function with missing return value, behavior at runtime .

As you are invoking undefined behaviour your display output is essentially meaningless.

edited May 23 '17 at 11:50

Community

1
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answered Dec 21 '13 at 14:27

mathematician1975

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Thats what I thought but it's not giving any error, even no compiler warnings. – Santosh Sahu Dec 21 '13 at 14:28
Have you requested ALL warnings from the compiler? – mathematician1975 Dec 21 '13 at 14:30
2

@SantoshSahu: The compiler is not obligated by the standard to issue any warnings for this, although it probably will if you enable them all as math suggested. – Benjamin Lindley Dec 21 '13 at 14:48

how exactly stack unwinding works between caller function and called function

1 Answers1