C programming

Question

#include<stdio.h>
#include<stdlib.h>
#include<conio.h>
int * show();
int main()
{
	int *p=show();
	//clrscr();
	printf("%d",*p);
}
int * show()
{
	int x=10;
	/*int *p;
	p=(int *)malloc(4);
	*p=x;*/
	return &x;
}

Here as the $show()$ ends variable $x$ gets destroyed but still I am able to access its value why is it so?

Comments

which compiler u r  using ? ..on gcc , it is showing segmentation fault which is correct

---

No bro it's working no error

Tried on gcc Linux and code blocks both its give only warning

Answer 1

#include<stdio.h>
int * show();
int main()
{
int *p=show();
printf("%d",*p); //Referring x that is de-allocated now but it's value is still exist on memory(in stack)
printf("%d",*p); //The memory (stack space) which is earlier allocated to show() is overwritten now by above printf so it will print a garbage value
return 0; 
}
int * show()
{
int x=10;
return &x;
} //e

Although scope and lifetime of variable x end as point e but its value is still there on the stack that's why 1st printf gives correct output. But after that, the memory is overwritten so 2nd printf will return a garbage value.

NOTE - GCC compiler returns 10 -2(this is a garbage value) with a Warning: function returns address of local variable.

Comments

@Soumya ,local variable x is stored in stack frame show(). So when it is deleted then why in main() , 1st printf() will give correct output.

I am getting segmentation fault because it is accessing memory which does not belongs to this process.

 

Answer 2

The code doesn't work and gives error in GCC, but works with a warning in ZAPCC. So the compiler you used to run this code is optimised to run through such errors by giving warning(s). Try your code here on different compilers, like here:

https://www.jdoodle.com/c-online-compiler

Comments

the variables are destroyed as the function ends but the value are constants and so it remains at that address

only and we are accessing it am I true?

Answer 3

@ankit I tired on code blocks and it is working fine there. Just giving a warning. 
Although this is the case of undefined behaviour. It's not always true that it will return the correct value. 

Check this modified code. 

#include<stdio.h>
int * show();
int main()
{
int *p=show();
printf("hello"); ​​​​​​
printf("%d",*p); \\memory is overwritten already. So it will print garbage value 
return 0; 
}
int * show()
{
int x=10;
return &x;
} //e

Comments

yes, but I don't know why it is showing segmentation fault on my machine. I have tried online gdb compiler, there it is also showing segmentation fault. I don't know what is the problem.

---

Because on your machine after the show() is destroyed, that memory is reused and allocated to printf() function that's why it is giving segmentation fault. But on my machine, although that memory is no longer belong to show() function, it is not cleaned up by the time printf() is executed so I am able to get its correct value. So it's completely machine dependent.

---

ok got it..thank you !