Question

Ie., why does the following work:

    char* char_array(size_t size){
            return new char[size];
    }

    int main(){
            const char* foo = "foo";
            size_t len = strlen(foo);
            char* bar=char_array(len);
            memset(bar, 0, len+1);
    }

But the following segfaults:

    void char_array(char* out, size_t size){
            out= new char[size];
    }

    int main(){
            const char* foo = "foo";
            size_t len = strlen(foo);
            char* bar;
            char_array(bar, len);
            memset(bar, 0, len+1);
    }

Answer 1

Passing ‘bar’ to char_array is passing a copy of the present value of that pointer at the time of the call – so ‘out’ in char_array points to the same thing as ‘bar’ but they’re completely isolated variables, and when char_array returns the newly allocated value is simply lost.

If you want to actually modify the ‘bar’ variable, you need to pass a pointer or reference to the bar variable itself, i.e.

void char_array(char** out, size_t size) {
    *out = new char[size];
}

...
char_array(&bar, len);

or

void char_array(char*& out, size_t size) {
    out = new char[size];
}

...
char_array(bar, len);
...