c语言指针错误 Access violation writing location 0x0000000000000000.

Question

我创了一个测试function能不能成功的case但当我运行的时候会给我报错

这个是公式
#include "number.h"
#define arraylength  5


int Printoutarray1(int* array1[])
{
    printf("their is your array1：\n");
    for (int i = 0; i < arraylength; i++)
    {
        printf("%d ", &array1[i]);
    }
    printf("\n");

    return Printoutarray1;
}

int Printoutarray2(int* array2[])
{
    printf("their is your array1：\n");
    for (int i = 0; i < arraylength; i++)
    {
        printf("%d ", &array2[i]);
    }
    printf("\n");

    return Printoutarray2;
}


int Mutiplyarrays(int array1[], int array2[], int sum[])
{
    for (int i = 0; i < arraylength; i++)
        // make the formular
        sum[i] = array1[i] * array2[i];
    

    return Mutiplyarrays;
```c

这是测试function能不能成功的程序
#include "../item3/number.h"
#define _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
int main(void)
{
    int expect1[] = { -1, 0, 20, 10, -24 };
    int expect2[] = { 1, 0, 20, 10, 24 };
    int sum = 0;
    int array1[] = { -1, 0, 5, 5, 3 };
    int array2[] = { 1,2,4,2,-8 };
    
    


    printf("TEST 1: ");
    sum = Mutiplyarrays(array1, array2, sum);
    if (sum == expect1[4])
        printf("pass\n");
    else
        printf("fail\n");

    
    printf("TEST 2: ");
    sum = Mutiplyarrays(array1, array2, sum);
    if (sum == expect2[4])
        printf("pass\n");
    else
        printf("fail\n");


    return sum;
}

这是给我报错的信息

Answer 1

代码里语法错误，见注释，供参考：

#include <stdio.h>
// 这个是公式
#include "number.h"
#define arraylength  5

int Printoutarray1(int* array1[])
{
    printf("their is your array1：\n");
    for (int i = 0; i < arraylength; i++)
    {
        printf("%d ", &array1[i]);
    }
    printf("\n");

    //return Printoutarray1;//这是要返回函数名？
    return 1;
}

int Printoutarray2(int* array2[])
{
    printf("their is your array1：\n");
    for (int i = 0; i < arraylength; i++)
    {
        printf("%d ", &array2[i]);
    }
    printf("\n");

    //return Printoutarray2;//这是要返回函数名？
    return 1;
}


int Mutiplyarrays(int array1[], int array2[], int sum)//int sum[] 形参是个数组 ？
{
    for (int i = 0; i < arraylength; i++)
        // make the formular
        sum = array1[i] * array2[i];  //sum[i]
    //return Mutiplyarrays; //这是要返回函数名？
    return sum;
}

//这是测试function能不能成功的程序
#include "../item3/number.h"
#define _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
int main(void)
{
    int expect1[] = { -1, 0, 20, 10, -24 };
    int expect2[] = { 1, 0, 20, 10, 24 };
    int sum = 0;
    int array1[] = { -1, 0, 5, 5, 3 };
    int array2[] = { 1,2,4,2,-8 };
    
    

 
    printf("TEST 1: ");
    sum = Mutiplyarrays(array1, array2, sum);//实参sum是个变量？
    if (sum == expect1[4])
        printf("pass\n");
    else
        printf("fail\n");
 

    printf("TEST 2: ");
    sum = Mutiplyarrays(array1, array2, sum);//实参sum是个变量？
    if (sum == expect2[4])
        printf("pass\n");
    else
        printf("fail\n");

    return sum;
}