传递指向 int 数组的指针，作为参数 C++

Question

I know even if i pass an array by typing arrayname as argument (ex: getArrayInput(arrayexample); ), it will copy only the adress value of first element not entire array,still i wonder why these code gives error. I know this not the way how it should implemented but i want to understand this error.

main.cpp|13|error: cannot convert 'int*' to 'int**' for argument '1' to 'void getArrayInput(int**)'|

#include <iostream>

using namespace std;
void getArrayInput(int * []);

int main()
{
    cout<<"Enter scores on by one.." << endl;
    cout<<"To terminate input enter -1"<<endl;

    int listof[10]={};
    int *ptScores =listof;
    getArrayInput(ptScores);




    return 0;
}

void getArrayInput(int * []){
    for(int i=0;i<10;i++){
        cin>>*(pt+i);
        if(*(pt+i))=-1){
            break;
        }
        else{
            cout<<"Enter next.."<<endl;
        }
    }
}

Answer 1

It is because

int * 

and

int[] 

are both of type

int *

therefore, you are here asking for a

int **.

try replacing

void getArrayInput(int * []) by     void getArrayInput(int *)

Answer 2

In C, arrays decay in to pointers. In some cases, they are interchangable.

ptScores is of type int* (pointer to int). getArrayInput expects an int*[] (array of pointers to int). int*[] decays in to int** (pointer to pointer to int).

The error says you're giving an int* (ptScores) to something that expects an int** (getArrayInput).

---

How do you fix this? Take an int* .

void getArrayInput(int* pt){
    for(int i=0;i<10;i++){
        cin>>pt[i];
        if(pt[i]=-1){
            break;
        }
        else{
            cout<<"Enter next.."<<endl;
        }
    }
}