array has no values?

[adam2016]

hey guys,

I declared a function that returns a pointer to an array,but when I try to print the values out in main all the values seem to be null or blank

what am I doing wrong?

thanks

  string* newArray(){

	string *ray = new string[4];
	for(int i = 0; i < sizeof(ray) / sizeof(string); i++){

		ray[i] = ("hello");

	}
    return ray;
}

main

string* arr = newArray();

    for(int i = 0; i < 4; i++){

    	cout << "array = " << arr[i] << endl;

    }

delete [] ray

[Repeater]

sizeof(ray)
Ray is a pointer, so probably 4 (or maybe 8). Nothing to do with the size of what ray is pointing to.

sizeof(string)
Who knows. 32 on one test system I tried.

So your code says:

for(int i = 0; i <4 / 32; i++){

		ray[i] = ("hello");

	}

What is 4/32 ? Well, that's an int divided by an int, so it's an int. So zero. So your code says:

for(int i = 0; i <0; i++){

		ray[i] = ("hello");

	}

So you never write anything into the strings.

[Cubbi]

as pointed out already, sizeof is the wrong thing to use. The "sizeof(ray) / sizeof(string)" C-style trick is for arrays, while what you have there is a pointer. It's important to understand the difference between arrays and pointers, especially if you plan on working with C.

In general, use of pointers and especially use of new and delete is a sign of bad C++ code: this is how the same task could be done today:

#include <vector>
#include <string>
#include <iostream>
std::vector<std::string> newArray() {
    return {4, "hello"};
}
int main() {
    auto arr = newArray();
    for(auto& s: arr)
        std::cout << "array = " << s << '\n';
}

live demo: https://wandbox.org/permlink/iX6SbbKRPHtqhaIJ

[adam2016]

that makes sense thanks guys