Pass array of struct pointers to function

Question

Trying to pass array of struct pointers to function doIt() . Looks my way is not correct since I can't get right second array element:

struct c {
    int a;
    char* b;
};


struct cc {
    int a;
    c* b;
};

char a[] = "aaa";
char b[] = "bbb";
char e[] = "eee";

c d1 = {1,a};
c d2 = {2,b};
c d3 = { 12,e };

cc g1 = { 123, &d1 };
cc g2 = { 321, &d2 };
cc g3 = { 333, &d3 };



void doIt( c *  s)
{
    cout << s->b;
    s++;
    cout << s->b;
}

What is right way to pass array of struct pointers?

please show examples of code that calls the doIT function. Also, it would be less confusing if you didn't reuse the same names (a, b) for multiple objects. — Scott Hutchinson, Sep 05 '21 at 16:41
Possible duplicate of https://stackoverflow.com/questions/1719051/array-of-pointers-as-function-parameter — Scott Hutchinson, Sep 05 '21 at 16:46
As shown, `doIt` doesn't take an array of pointers. It takes a pointer to an array of objects of type `c`. The rest of the code doesn't declare an array of `c` suitable for passing to this function. It's unclear what you are trying to achieve. — Igor Tandetnik, Sep 05 '21 at 16:53

score 0 · Answer 1 · answered Sep 05 '21 at 17:04

Raw arrays in C (and C++) are just pointers. They point to the first element of an array. For example, if you want an array of int, you would write it like int* array. If you want an array of struct c, you would write it like c* array. If you want an array of pointers to struct c, you would write it like c** array.

To access elements, don't use array++, use array[i] where i is the index (position) of the element you want to access, 0 being the index of the first element, 1 the second, etc.

So, your code should look like this:

void doIt(c** s)
{
    cout << s[0]->b; // s[0] is the first element
    cout << s[1]->b; // s[1] is the second
}

Note that in C++, it is preferred to use std::vector instead of raw arrays.

void doIt(std::vector<c*> s)
{
    cout << s[0]->b;
    cout << s[1]->b;
}

score 0 · Answer 2 · answered Sep 05 '21 at 17:24

If you want to pass array to a function you need also pass length of this array:

#include <iostream>
#include <vector>

struct c {
    int a;
    char* b = nullptr;
    size_t size = 0;
};

void doIt(c* all, size_t length);

int main()
{
    char a[] = "aaa";
    const size_t sizeOfA = sizeof(a)/sizeof(a[0]);
    char b[] = "bbb";
    const size_t sizeOfB = sizeof(b)/sizeof(b[0]);
    char e[] = "eee";
    const size_t sizeOfE = sizeof(e)/sizeof(e[0]);

    c d1 {1, a, sizeOfA}; 
    c d2 {2, b, sizeOfB};
    c d3 {12, e, sizeOfE};

    c all[] = {d1, d2, d3};
    const size_t length = sizeof(all)/sizeof(all[0]);

    doIt(all, length);

return 0;
}

void doIt(c* all, size_t length)
{
    if (!all)
    {
        std::cerr << "Pointer to array is null" << std::endl;
    }       
    for (size_t i = 0; i < length; ++i)
    {
        for (size_t j = 0; j < all[i].size; ++j)
        {
            std::cout << all[i].b[j];
        }
        std::cout << std::endl;
    }
}

You can use std::vector. So, you don't need to use adittional argument (length of the vector):

#include <iostream>
#include <vector>
#include <string>

struct c {
    int a;
    std::string b;
};

void doIt(const std::vector<c>& myVector);

int main()
{
    std::vector<c> myVector;
    myVector.emplace_back(1, "aaa");
    myVector.emplace_back(2, "bbb");
    myVector.emplace_back(12, "eee");

    doIt(myVector);

return 0;
}

void doIt(const std::vector<c>& myVector)
{
    for (size_t i = 0; i < myVector.size(); ++i)
    {
        std::cout << myVector[i].b << std::endl;
    }
}

Pass array of struct pointers to function

2 Answers2