What is the reason for the following output in C get changed dynamically

Question

Here is the code segment I am dealing with. It outputs 500, but I can not figure out, why this is happening.

#include<stdio.h>

int *f(int x){
    int p;
    p = x;
    printf("p is : %d\n",p);
    return &p;
}

int *g(int x){
    int y;
    y = x;
    printf("p is : %d\n",y);
    return &y;
}

int main(){
    int *x,*y;
    x = f(1000);
    y = g(250);
    *x = *x +250;
    printf("%d %d\n",*x,*y);

    return 0;
}

What is the reason for the output of 500?

You're using an address of a variable that has gone out of scope. — Fiddling Bits, Feb 12 '20 at 15:44
Doesn't our compiler warn you about returning stack pointers? — Blaze, Feb 12 '20 at 15:45
@Blaze the program is not returning stack pointers, it's returning pointers to local variables. — Jabberwocky, Feb 12 '20 at 15:48
@FiddlingBits: The applicable concept is lifetime or storage duration, not scope. — Eric Postpischil, Feb 12 '20 at 15:49
I've closed as duplicate of [this question](https://stackoverflow.com/questions/6441218/can-a-local-variables-memory-be-accessed-outside-its-scope). Although this question is about C++ the problem is _exactly the same_. — Jabberwocky, Feb 12 '20 at 15:54

score 3 · Answer 1 · answered Feb 12 '20 at 15:45

You're returning the address of a local variable.

The lifetime of the variables p and y in the functions f and g ends when those functions return. As a result, the pointers to them become invalid and attempting to dereference them results in undefined behavior.

Rather than returning pointers, just return the value directly.

some user · Accepted Answer · 2020-02-12T15:56:57.533

Look at your function:

int *f(int x){
    int p; 
    p = x;
    printf("p is : %d\n",p);
    return &p;
}

warning: address of local variable 'p' returned [-Wreturn-local-addr]

Here variable p is created in the scope of f() after the return of function call this functions stack frame get destroyed so are everything in it, thus unexpected results. You can either return the value directly or dynamically allocate memory using malloc(), in this case variable will be stored in heap instead of f() or particular functions stack later on you can free the memory when you're done.

What is the reason for the following output in C get changed dynamically

2 Answers2