Question

Possible Duplicate:
Can a local variable’s memory be accessed outside its scope?

#include <iostream>
#include <string>

using namespace std;

char* reverse(char* input) 
{
    char b[11];

    for (int i=0; i<11; i++)    {b[10-i]=input[i];}
    for (int u=0; u<11; u++)    {cout<<b[u];}
    cout<<endl;
    return &b[0];
}

int main ()
{
     char ali[]="ali is good";
     char *a=&ali[0];
     char *b=reverse(a);

     for (int i=0; i<11; b++, i++)  
        {
            cout<<*b+i<<endl;
        }
     cout<<endl;
     system("pause");
     return 0;
}

This is a simple program to reverse a string, it works well when you print the result in the function, but when passing the pointer of the string to Main then printing it, it prints garbage-like things.

Why is it strange? because if I print *b+2 or *b+5 or whatever const I like, it works well, but if I replace the const with an int in a loop like *b+i then it does not work!!

P.S. this is not a HW!! I “was” a longtime programmer, shame on me for forgetting all of this.

Any ideas?

Answer 1

The variables declared inside a function are automatic storage duration objects and go ou tof scope at the end of a function (and thus are no longer valid). To allocate storage that lives longer than a function you need dynamic storage duration objects (these are allocated via new):

char *b = new char[11];