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

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.