是否可以将指针数组作为“ int”参数传递给函数，从而将地址转换为整数？

Question

Beginner c++ programmer, writing code to solve Sudoku puzzles. I'll keep the background info short to avoid confusion.

I have an array of pointers, int *P[9] , I have assigned each entry a specific address. I want to assign these addresses to another array of pointers, int *B[81] .

P[0] should correspond to B[0] , P[1] to B[8] , and so on.

When I pass these to a function:

void (int B[  ], int P[  ] ) {...}

it seems like the function is converting the address P[ ] is pointing to into an integer value. Before the function is called P[0] points to the address 0x7fff978d46b0 , if I check the value of P[0] inside the function it's a number like `48782346 .

#include<iostream>
using namespace std;

void assign_box(int matrix[], int P[])
{
    cout << "P[0] in function: " << P[0] << "\n";
    matrix[0]=P[0];
}

int main()
{
    int table[9][9];
    //Initialise table entries to 0
    for(int i=0; i<9; i++)
    {
        for(int j=0; j<9; j++)
        {
            table[i][j]=0;
        }
    }
    //Assign addresses to vector P, for brevity P is of length one
    int *P[1];
    P[0]=&table[0][0];

    cout<< "P[0] before function: " << P[0] << "\n";

    int*B[81];
    assign_box(B[81], P[9]);
}

If it did this and worked I wouldn't care, but unfortunately when I assign B[0] = P[0] , it hits me with a Segmentation fault (core dumped) , which makes me wonder is the function trying to assign the pointer B[0] to the address 48782346.

Is it possible for the function to convert an address into an integer value?

Apologies if my question is unclear or verbose, first time asker. And thank you for edits.

Answer 1

If you dereference int*[] (or int** ), you get an int* . If you dereference an int* , you get an int . This is exactly what you are doing, and why you end up with an int at the end.

//main
int *P[1]; //Array of pointers to int
int *B[81]; //Array of pointer to int
assign_box(B[81], P[9]); //Pass in two pointers to int

//assign_box
matrix[0]=P[0]; //assign int to int

You probably meant to call assign_box like assign_box(B, P) , and have the signature be void assign_box(int *B[], int *P[]); . This would then allow you to assign one pointer inside an array to another pointer inside an array.

There are multiple things that could be causing segmentation faults, but they all stem from invalid array indices. If an array is declared like type identifier[size]; , it has valid indices from 0 to size - 1 . So, int *B[81]; means B[81] is invalid.

Answer 2

You're passing in the wrong parameters. You're trying to pass in an array object B[81] which does NOT EXIST. You only have B[0] - B[80]. Also, B[80] isn't an int pointer. It's an int within an int array. P[9] is a pointer to an array of integers. So, you're trying to pass an integer in an array slot that does not exist into a parameter that does not take integers -- it takes integer arrays.

#include<iostream>
using namespace std;

void assign_box(int matrix[], int P[])
{
    cout << "P[0] in function: " << P[0] << "\n";
    matrix[0]=P[0];
}

int main()
{
    int table[9][9];
    //Initialise table entries to 0
    for(int i=0; i<9; i++)
    {
        for(int j=0; j<9; j++)
        {
            table[i][j]=0;
        }
    }
    //Assign addresses to vector P, for brevity P is of length one
    int *P[1];
    P[0]=&table[0][0];

    cout<< "P[0] before function: " << P[0] << "\n";

    int*B[81];
    assign_box(B[81], P[9]); // WRONG

}