通过作为C ++函数的指针传递来访问std :: array元素的正确方法是什么？

Question

The code snippet is as follows. The error is in foo function for the cout line:

typedef struct Datatype {
    int first;
    int second;
} Datatype;

void foo(std::array<Datatype, 100>* integerarray){
    cout << *integerarray[0].first << endl; //ERROR: has no member first
}

void main() {
    std::array<Datatype, 100> newarray;
    for(int i=0; i<100; i++)
        newarray[i] = i;
    }
    foo(&newarray);
}

Answer 1

Because of operator precedence, *integerarray[0].first is translated as *(integerarray[0].first) , which is not what you want. You need to use (*integerarray)[0].first .

cout << (*integerarray)[0].first << endl;

You can make your life simpler by passing a reference.

void foo(std::array<Datatype, 100>& integerarray){
  cout << integerarray[0].first << endl;
}

Also, you don't need to use typedef struct DataType { ... } DataType; in C++. You can use just struct DataType { ... };

Answer 2

newarray[i] = i;

In this line you have missed adding value to structure variables. Also you have passed array as a reference to function. Removing it and passing just the name of array will pass base address of array. I am adding following code for your reference:

#include<iostream>
#include <array>
struct Datatype{
int first;
int second;
}

typedef Datatype varInts;
void display(std::array<varInts,20> &dummy)
{
int b =5;
for(int i=0; i<20; i++)
{
dummy[i].first =b++;
dummy[i].second = b+5; //Give any logic you wish.just adding different values;
b++;
}
}
int main()
{
std::array<varInts,20> data;
int a =1;
for(int i=0;i<20;i++)
{
 data[i].first = a++;
data[i].second = a+5;
a++; //Just adding values for example
}
display(data);
return 0;
}

It runs without error.Hope it helps!!