Why can't I pass string* instead of string[] as argument?

Question

int main(int args, char* argv[]) {
    string* list;
    listDevices(list);
    printf("device 0: %s\n", list[0]); // prints some junks
}

void listDevices(string* list) {
    list = new string[1];
    list[0] = "abc";
    printf("device 0: %s\n", list[0]); //prints "abc"
}

In the above code, list array is initialized and assigned some values in the listDevices method, but when I print it outside the function, some junks would be printed.

Why does it work correctly if I send string list[1] as input to the above method but not when sending string* list (although both are pointers)?

`string list[1]` is an array, not a pointer. In "the above code", `list` is a pointer, not an array. — molbdnilo, Oct 12 '15 at 08:59
This is a very messy code. At least use `delete` when you use `new`. — Lukáš Bednařík, Oct 12 '15 at 09:01

Baum mit Augen · Answer 1 · 2015-10-12T09:03:23.007

You only change a local copy of the pointer in listDevices. The pointer in main remains unchanged. You probably want to return a pointer to the new list:

string *listDevices() {
    string *list = new string[1];
    list[0] = "abc";
    printf("device 0: %s\n", list[0]); //prints "abc"
    return list;
}

int main(int args, char* argv[]) {
    string *list = listDevices();
    printf("device 0: %s\n", list[0]); // Will now work
}

Alternatively, you can pass the pointer by reference, but I prefer the return method for stylistic reasons.

By the way, you probably do not need to use a pointer and printf in the first place. Most likely, your code should look somewhat like this:

std::string listDevices () {
    std::string ret = "abc";
    std::cout << ret << "\n";
    return ret;
}

int main () {
    std::string str = listDevices();
    std::cout << ret << "\n";
}

can not pass the argument here: string *list = listDevices(list); — vishal, Oct 12 '15 at 09:02

score 2 · Accepted Answer · answered Oct 12 '15 at 08:59

2

list is a variable of type pointer-to-string, which you pass by value. Changing list itself inside the method won't carry the changes outside the scope of the method.

You want to pass it by reference:

void listDevices(string*& list)

answered Oct 12 '15 at 08:59

Luchian Grigore

253,575
64
457
625

score 1 · Answer 3 · edited May 23 '17 at 11:58

You need to pass list as reference to function as:

void listDevices(string* &list) {
    list = new string[3];
    list[0] = "abc";
    printf("device 0: %s\n", list[0].c_str()); //prints "abc"
}

int main(int args, char* argv[]) {
    string* list=nullptr;
    listDevices(list);
    printf("device 0: %s\n", list[0].c_str()); //now prints "abc"
    // Don't forget to do `delete[] list` at the end !
    return 0;
}

Note: As you tag c++ in your question, check this link which indicate to using .c_str() while printing a std::string with printf.

caution: Don't forget to do delete[] list at the end !

Why can't I pass string* instead of string[] as argument?

3 Answers3