2

We currently learn about pointers in combination with arrays in C/C++ and have to implement some function, for which we have a given function call. The objective is to print out the array values by calling the function with the array adress. The output is not what was expected.

I tried printing the the array values with & and * as well as with none of those.

void ausgabe1D_A(double (*ptr1D)[4]){
    int i{0};
    for (i = 0; i < 4; i++)
        std::cout << *ptr1D[i] << "  ";
}

int main()
{
    double ary1D[4] = {1.1, 2.2, 3.3, 4.4};

    ausgabe1D_A(&ary1D);
}

I expected the output to be:
1.1 2.2 3.3 4.4

Instead I got:
0x61fdf0 0x61fe10 0x61fe30 0x61fe50 (with and without &)
1.1 9.09539e-318 0 0 (with *)

EDIT: Sorry if that wasn't clear, but we have to call the function with &ary1D. We are trying out different ways and your ways are coming up in the exercise in following functions but for now we need it with the &-Operator.

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4 Answers 4

Reset to default
2

Although others provide a possible solution, I will try to explain why your code fails. Expression double(*ptr1D)[4] refers to a pointer to an array of four elements (doubles). To get values, you should firstly dereference the pointer and only then print out its n-th value, as follows:

void ausgabe1D_A(double (*ptr1D)[4])
{
  for (int i = 0; i < 4; i++) {
    std::cout << (*ptr1D)[i] << " "; // <-- note the parenthesis
  }
}
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3 Comments

Thank you very much, this answer is what I've been looking for! Did i get it right, that I had a pointer pointing on an array which essentialy is a pointer and that's why the first value of the array was printed correctly while the others were not?
@xCuse The problem was probably that *ptr1D[i] points to base of array + i * length of whole array. Hence, the out-of-bound accesses. Demo on coliru
An array decays to a pointer to the first element. So ary1D decays to a pointer to a double. Going the other ways double (*ptr1D)[4] is what a two dimensional array decays to. Your input (with the & operator) is a 1x4 two dimensional array. *(ptr1D[i]) [parents for emphasis] then prints the first double of each row. Since your input only has one row the rest are garbage, out of bounds array access.
2

When you reference an array by its name, you are referencing it by its pointer, so you function argument is wrong. It should be like

void ausgabe1D_A(double *ptr1D, size_t size){
    int i{0};
    for (i = 0; i < size; i++)
        std::cout << ptr1D[i] << "  ";
}

int main()
{
    double ary1D[4] = {1.1, 2.2, 3.3, 4.4};

    ausgabe1D_A(ary1D,4);
}

Here is a picture of the memory of a short integer array for example

enter image description here

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4 Comments

Hey, I'm sorry that i didn't make it clear but I need to call the function with &ptr1D
@XCuse, no! ary1D is a pointer. it's like &(ary1D[0]). Arrays are pointers to an array of the type.
Thank you, we already learned that an array is nothing else than a pointer to an adress where the array element 0 is saved in and that ary[i] is equal to *(ary+i). Nonetheless I had to use the given parameter &ary1D. And sorry it shoudl have been &ary1D not ptr1D.
If using &ary1D is mandatory, then you can reference it like (*ary1D)[i].
0

You can take advantage of the fact that arrays can decay into pointers, so you'd write something like

#include <iostream>

void ausgabe1D_A(double* ptr1D){
    for (std::size_t i = 0; i < 4; i++)
        std::cout << ptr1D[i] << "  ";
}

int main()
{
    double ary1D[4] = {1.1, 2.2, 3.3, 4.4};
    ausgabe1D_A(ary1D);
}

The way this works is that the pointer ptr1d is essentially the address of the first element of the array.

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1 Comment

Hey thank you for your answer! Unfortunately we have to call the function like this ausgabe1D_A(&ary1D);
0

I think you are looking for way to pass using & operator. Here you go.

void print( int (*arr)[4])
{
    int *d = *arr;
    for(int i = 0 ;i <4 ; ++i)
    {
        std::cout<<d[i]<<std::endl;
    }
}
int main()
{
    int arr[] = {1,2,3,4};
    print(&arr);

}

Let me know if you found it hard to understand.

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1 Comment

Thank you, this helps a lot!

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