Why cout a simple char* is not working?

Question

I want you to look at the following two codes and tell me why the first one works and the other one doesn't.

First Code (Working):

#include <iostream>
#include <string.h>

using namespace std;

char* getCoordinates(char* AndroidID)
{//Access the database and get the South Latitude value
    //This is here is dummy data for testing
    const char* S = "27.19122859";
    const char* N = "27.19245011";
    const char* W = "31.17657602";
    const char* E = "31.17657602";
    char X[100]="";
    strncat(X, "S", 1);
    strncat(X, S, strlen(S));
    strncat(X, "N", 1);
    strncat(X, N, strlen(N));
    strncat(X, "W", 1);
    strncat(X, W, strlen(W));
    strncat(X, "E", 1);
    strncat(X, E, strlen(E));
    char* Y = X;
    cout<<Y;
    return Y;
}
int main()
{
    char* sda=NULL;
    getCoordinates(sda);
    return 0;
}

This prints out correctly the string I want to compose

S27.19122859N27.19245011W31.17657602E31.17657602

But when I try to do the same thing but print out the return value in main() things get a little weird Second Code (Not working)

#include <iostream>
#include <string.h>

using namespace std;

char* getCoordinates(char* AndroidID)
{//Access the database and get the South Latitude value
    //This is here is dummy data for testing
    const char* S = "27.19122859";
    const char* N = "27.19245011";
    const char* W = "31.17657602";
    const char* E = "31.17657602";
    char X[100]="";
    strncat(X, "S", 1);
    strncat(X, S, strlen(S));
    strncat(X, "N", 1);
    strncat(X, N, strlen(N));
    strncat(X, "W", 1);
    strncat(X, W, strlen(W));
    strncat(X, "E", 1);
    strncat(X, E, strlen(E));
    char* Y = X;
    return Y;
}

int main()
{
    char* sda=NULL;
    char* T=getCoordinates(sda);
    cout<<T;
    return 0;
}

This outputs:

S27. N27.19245011W31.`%@ \╞G ░■) $âG @■)

What is the problem here?

Why aren't you using `std::string`, which would just work here? — Alan Stokes, May 29 '16 at 05:55
[The compiler could have solved this.](http://coliru.stacked-crooked.com/a/69795df051f4d148) — chris, May 29 '16 at 05:57

score 1 · Answer 1 · answered May 29 '16 at 05:53

1

You are returning a pointer to an array of char which has lifetime only as long as the function. When the function returns the array is no longer valid.

answered May 29 '16 at 05:53

John Zwinck

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Thanks :) I'll try to do it with a string then convert that string to a char* that is needed as an input to another function. – Samer Bastawy May 29 '16 at 05:59
@SamerBastawy You can always use `c_str()` from a `std::string`. – πάντα ῥεῖ May 29 '16 at 06:01

score 1 · Answer 2 · answered May 29 '16 at 05:53

1

It dos not work because your function getCoordinates return pointer to buffer which has local scope of function getCoordinates and gets destroyed after the functions returns. Use std::string instead of char and it will be working.

answered May 29 '16 at 05:53

Tomas Kubes

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Thank you. Appreciate it – Samer Bastawy May 29 '16 at 06:00

Why cout a simple char* is not working?

2 Answers2