Pointer to array in function in c++

Question

I am a beginner to c++. Pointer is quite confusing to me. Especially on how to use it in functions and array. I tried to create a pointer to array in function and just output it. However it keeps giving me the address of the array instead of the value.

void testing(int* arr){
    cout << arr << endl;
}

int main()
{
    int my_arr[]{ 4,7,1 };
    testing(my_arr);

    string y;
    getline(cin, y);
    return 0;
}

I tried using testing(&my_arr); to output value but it give me errors:

argument of type "int (*)[3]" is incompatible with parameter of type "int *

'void testing(int *)': cannot convert argument 1 from 'int (*)[3]' to 'int *'

Thanks a lot for any help!

The error tells you everything you need to know. An `int*` is not a pointer to an int array. — Charles, Sep 15 '17 at 02:59
Are you sure you aren't calling the function with `testing(&my_arr);`? That's what the error suggests to me. — Ken Y-N, Sep 15 '17 at 03:03
@KenY-N you're saying that an `int*` is the same as a *pointer* to an int array? You're mistaken. A pointer to an int array is `int**` or `int *[]` as shown in the error. — Charles, Sep 15 '17 at 03:12
"it keeps giving me the address of the array instead of the value". That's because the address (of the first item, which is the same as the address of the array) is what you pass to the stream. It outputs exactly what you tell it to output. To output the array values use a loop. It will need to know the number of items. — Cheers and hth. - Alf, Sep 15 '17 at 03:12
@KenY-N, yes for the errors i use `testing(&my_arr);` instead because `testing(my_arr);` gives me address. — skyzhuzhu, Sep 15 '17 at 03:12
OP, if you want to print the values in an array you need to loop to print them out. `std::cout` won't do that for you — Charles, Sep 15 '17 at 03:13
@Charles, how to create pointer to array? I see tutorials they teach int* arr — skyzhuzhu, Sep 15 '17 at 03:15
@engkhsky `int*` is ___not___ a pointer to an array! It's just a pointer to an `int`. — Charles, Sep 15 '17 at 03:20
Read about [decaying](https://stackoverflow.com/questions/1461432/what-is-array-decaying) — Passer By, Sep 15 '17 at 03:23
There's no built-in function for outputting all of the members of a container (except string...) , you will need to make your own — M.M, Sep 15 '17 at 05:23

Charles · Answer 1 · 2017-09-15T03:24:48.617

To print the values in an array rather than the starting address, you need to use a loop.

#include <iostream>
#include <string>

// note extra param for length of array.
void testing(int* arr, int len){
    for (int i = 0; i < len; ++i)
        std::cout << arr[i] << " ";
    std::cout << "\n";
}

int main()
{
    int my_arr[]{ 4,7,1 };
    testing(my_arr, 3);

    return 0;
}

You can't pass testing(&my_arr) because &my_arr is of type int (*)[] as per the error message you received. That is not the same as int*.

score 0 · Accepted Answer · answered Sep 15 '17 at 05:12

0

for printing the arrays, you can either use the array index or pointers arithmetic. The test function could also be written as

void testing(int* arr, int len) {
    for (int ctr = 0; ctr < len; ctr++) {
        std::cout << *(arr + ctr) << std::endl;
    }
} 

int main()
{
    int my_arr[]{ 4,7,1 };
    testing(my_arr, 3);

    return 0;
}

answered Sep 15 '17 at 05:12

Daksh Gupta

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[No need for std::endl](https://stackoverflow.com/questions/35580919/should-stdendl-always-be-used) – Charles Sep 15 '17 at 16:16

score 0 · Answer 3 · answered Sep 15 '17 at 07:22

In testing() you are trying to use arr element without its index. Here arr is the only base memory address of that memory. To get value from there you have to specify index.

void testing(int* arr, int len)
{
    for(int i = 0; i < len; i++)
    {
        cout << arr[i] << endl;
    }
}

In main() you can pass a length of an array.

int main()
{
    int my_arr[]{ 4,7,1 };
    testing(my_arr, sizeof(my_arr) / sizeof(int));
    return 0;  
}

Pointer to array in function in c++

3 Answers3