C-返回局部指针与局部变量

Question

My question is regarding these two cases:

#include <stdio.h>
int *foo1();
int *foo2();

int main()
{
   printf("so it's %d\n",*foo1());
   printf("so it's %d\n",*foo2());
}

int *foo1()
{
   int i1 = 5;
   return &i1;
}

int *foo2()
{
   int i2 = 5;
   int *p = NULL;
   p = &i2;
   return p;
}

case1: When its the case with foo1(), we get an error because we are trying to return a copy of address to main whose data has been deleted(when we exit foo1() function)

case2: But in foo2() , it doesn't give an error though we are returning a copy to a pointer of local variable whose data will be deleted after we exit the foo2() function, why is it so?

TL;DR: why foo2() doesn't give an error but foo1() does?

TIA.

Answer 1

The behaviour of both functions foo1 and foo2 are undefined , in both C and C++.

You are not allowed to dereference a pointer that points to a variable with automatic storage duration that's no longer in scope.

Tomorrow, foo2() may well give you an error too. Or the compiler might eat your cat.

Answer 2

In both cases, you are invoking undefined behavior by returning a pointer to a local variable and dereferencing it.

Invoking undefined behavior does not mean you will always crash. It means the behavior of the program is undefined. It could crash, it could output strange results, or it could appear to work properly. As you have seen, this behavior manifests itself in two of these ways in your program.

Making a seemingly unrelated change, such as adding an unused local variable or a printf for debugging, can change the way undefined behavior manifests itself.

Answer 3

Both are undefined behavior, you cannot rely on the compiler giving you warnings for undefined behavior.

You are essentially doing the same thing in both functions and in fact both might result in the same assembly code to be generated.

Answer 4

case1和case2的行为均未定义，您无法将指针返回给局部变量