CODEWITHSUNDEEP
c++stringfunctionlpcstr
Giortmants Romskij
Giortmants Romskij

Reputation: 3

The return value of a LPCSTR function becomes trashy

I was coding a simple XOR encryption program and then I noticed that the return value of a function is not the kind I expect to see.

I just can't find a problem in my code, could anyone help me?

Here's the program code:

#include "windows.h"
#include "iostream"

using namespace std;


LPCSTR hasala(string original, char key){

  string changed;

  for (int temp = 0; temp < original.size(); temp++){
  changed += original[temp] ^ (int(key) + temp) % 255;
  }

  cout << changed.c_str()<<"\n\n";//works, output "acagaca"

  LPCSTR adart = changed.c_str();

  cout << adart<<"\n\n";//works, output "acagaca"

  return adart;

}

int main(){

    cout << hasala("abcdefg", 0);//doesn't work, output "||||@ER|||"...

    cout << "\n\n";

    Sleep(8000);

    return 0;
}

Upvotes: 0

Views: 2080

Answers (3)

nos
nos

Reputation: 229274

You're returning changed.c_str() , which is just a pointer into the buffer managed by changed. changed is a local variable, and is destroyed when the function ends, so you're ending up returning a pointer to something that no longer exists.

Return the std::string changed directly.

Upvotes: 3

Marius
Marius

Reputation: 837

you are returning a local variable allocated on stack.

basically your code translate to:

LPCSTR hasala(string original, char key){
string changed;
return changed.c_str();
}

either allocate the memory (caller must free it) or return a std::string

Upvotes: 1

Puppy
Puppy

Reputation: 147018

The pointer that you have is to memory owned by the std::string on the stack - when the std::string goes out of scope, it frees it, and then you read it. You must return std::string, not LPCSTR.

Upvotes: 3

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