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?

CodePudding user response：

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;
}

CodePudding user response：

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;
    }
}