Question

#include <iostream>
#include <string.h>
using namespace std;

void newBuffer(char* outBuffer, size_t sz) {
    outBuffer = new char[sz];
}

int main(void) {

    const char* abcd = "ABCD";
    char* foo;
    foo = NULL;
    size_t len = strlen(abcd);
    cout<<"Checkpoint 1"<<endl;
    newBuffer(foo, len);
    cout<<"Checkpoint 2"<<endl;

    cout<<"Checkpoint 2-A"<<endl;
    memset(foo, '-', len);
    cout<<"Checkpoint 3"<<endl;
    strncpy(foo, abcd, len);
    cout<<"Checkpoint 4"<<endl;
    cout << foo << endl;

    int hold;
    cin>>hold;
    return 0;

}

This program crashes between checkpoint 2-1 and 3. What it tries to do is to set the char array foo to the char ‘-‘, but it fails because of some access issues. I do not understand why this happens. Thank you very much in advance!

Answer 1

Your newBuffer function should accept the first parameter by reference so that changes made to it inside the function are visible to the caller:

void newBuffer(char*& outBuffer, size_t sz) {
    outBuffer = new char[sz];
}

As it is now, you assign the result of new char[sz] to the local variable outBuffer which is only a copy of the caller’s foo variable, so when the function returns it’s as if nothing ever happened (except you leaked memory).

Also you have a problem in that you are allocating the buffer to the size of the length of ABCD which is 4. That means you can hold up to 3 characters in that buffer because one is reserved for the NUL-terminator at the end. You need to add + 1 to the length somewhere (I would do it in the call to the function, not inside it, because newBuffer shouldn’t be specialised for C-strings). strncpy only NUL-terminates the buffer if the source string is short enough, so in this case you are only lucky that there happens to be a 0 in memory after your buffer you allocated.

Also don’t forget to delete[] foo in main after you’re done with it (although it doesn’t really matter for a program this size).