“int*”类型的参数与 C++ 中“int**”类型的参数不兼容错误

Question

我的教授要求我创建函数并使用指针 为每个函数传递一个整数数组作为参数，但是当我插入他创建的测试代码时，我得到了“int*”类型的参数与“int**”类型的参数不兼容的错误。

#include <iostream>
#include "homework.h"
using namespace std;


int main() {
int a[10] = { 3, 5, 6, 8, 12, 13, 16, 17, 18, 20 };
int b[6] = { 18, 16, 19, 3 ,14, 6 };
int c[5] = { 5, 2, 4, 3, 1 };

Homework h;
// testing initialize_array
h.print_array(a, 10); // ERROR
h.initialize_array(a, 10); // ERROR
h.print_array(a, 10); // ERROR

h.print_array(b, 6); // ERROR
h.selection_sort(b, 6); // ERROR
h.print_array(b, 6); // ERROR

cout << "Factorial of 5 = " << h.factorial (5) <<endl; //print: 120
    c[0] = h.factorial(c[0]);
c[1] = h.factorial(c[2]);
h.print_array(c, 5); // ERROR
return 0;
})

这是作业的源文件：

#include <iostream>
#include "homework.h"
using namespace std;

void Homework::initialize_array(int *numArr[], int size)
{

for (int count = 0; count < size; count++) {
    // if divisible by 2, replace the value to 0.
    if (count % 2 == 0) {
        *numArr[count] = 0;
    }
    else {
        *numArr[count] = 1;
    }
}
}

void Homework::print_array(int *numArr[], int size)
{
// prints each values in the array.
for (int count = 0; count < size; count++) {
    if (count == 0) {
        cout << *numArr[count];
    }
    else {
        cout << ", " << *numArr[count];
    }
}
cout << endl;
}

void Homework::selection_sort(int *numArr[], int size)
{
int i, j, minIndex;

// determine the minimum value.
for (i = size - 1; i > 0; i--) {
    minIndex = 0;
    for (j=1; j<=i; j++) {
        if (*numArr[j] < *numArr[minIndex])
            minIndex = j;
    }

    // swap values.
    int temp = *numArr[minIndex];
    *numArr[minIndex] = *numArr[i];
    *numArr[i] = temp;
}
}

 int Homework::factorial(int num)
{
 if (num == 0 || num == 1)
    return 1;
else
    // if not 0 or 1, recall the function.
    return(num * factorial(num - 1));
}

Answer 1

如果您的教授想显示指针，您可以这样做：

void Homework::print_array(int *numArr, int size)//exact same as int numArr[] as it decays to a pointer
{
    for (int count = 0; count < size; count++) {
    if (count == 0) {
        cout << numArr[count];
    }
    else {
        cout << ", " << numArr[count];
    }
}
cout << endl;
}

int main()
{
   int a[10] = { 3, 5, 6, 8, 12, 13, 16, 17, 18, 20 };

    Homework h;

    h.print_array(a, 10);

    return 0;
}

编辑：在想也许你的教授希望你演示一个这样的指针数组：

void Homework::print_array(int *numArr[], int size)
{
    for (int count = 0; count < size; count++) {
    if (count == 0) {
        cout << *numArr[count];
    }
    else {
        cout << ", " << *numArr[count];
    }
}
cout << endl;
}

int main()
{
   int a[10] = { 3, 5, 6, 8, 12, 13, 16, 17, 18, 20 };
   int* arrayPtr[10];
   for(int i=0; i<10; i++){
       arrayPtr[i] = &a[i];
   }    
    Homework h;   
    h.print_array(arrayPtr, 10);
    return 0;
}

Answer 2

你不应该有一个指针数组。你应该有一个整数数组int *numArr[]：

void Homework::print_array(int numArr[], int size) //remove the extra * on front
{
// prints each values in the array.
for (int count = 0; count < size; count++) {
    if (count == 0) {
        cout << numArr[count];
    }
    else {
        cout << ", " << numArr[count];
    }
}
cout << endl;
}

好吧，如果你必须使用int *numArr[]，那么保留原来的print_array函数，但是如果你尝试打印它，使用h.print_array(&a, 10);。