如何解释这个奇怪的输出？ 关于指针和临时变量

Question

Who knows why the output is such?
Although it's wrong to use a pointer like that, I would still like to understand why it behaves the way it does.

int* foo()
{
    int a=9;
    int *p=&a;
    //definitely not right to return a pointer to an invalid varible
    return p;
}
int main(int argc, char* argv[])
{
    int *p=foo();
    cout<<*p<<endl;//9
    cout<<*p<<endl;//2357228
    *p=2;
    cout<<*p<<endl;//2
    (*p)++;
    cout<<*p<<endl;//2357229
    cout<<*p<<endl;//2357228
    cout<<*p<<endl;//2357228
    (*p)++;
    cout<<*p<<endl;//2357229
    cout<<*p<<endl;//2357228

    return 0;

}

Answer 1

Returning a pointer/reference to a variable local to an function results in Undefined Behavior . A local/automatic variable is guaranteed to be alive and valid only in the scope( { , } ) in which it is defined not beyond that scope.

Undefined behavior means that the program can show any behavior and it is allowed to do so by the C/C++ standards.It is meaningless to try to find reasoning of observed behavior after Undefined behavior has occured because this behavior may/may not be consistent or cannot be relied upon.

On a bright side any good commercial compiler will provide you warning about such a code.

Answer 2

Although it's wrong to use a pointer like that, I would still like to understand why it behaves the way it does.

Because p is left pointing at a location in stack memory that keeps getting written over by cout's << method. Every time you use cout, the value of p may change.

Code like this is dangerous as it can corrupt the stack and cause your program to crash.

Answer 3

What you need to remember is that a and p are both scope level variables. Thus, returning the address of a variable that only exists in the stack (but is not dynamically allocated on the heap) is undefined behavior. Since we no longer know what gets written to there once it's left that address space. Ergo, returning a pointer to a local variable is undefined behavior.