如何声明指向int数组的指针数组

How to declare an array of pointers to int arrays?

本文关键字：数组 int 指针 何声明 声明      更新时间：2023-10-16

我试图声明一个指针数组，每个指针指向不同大小的int数组。有什么想法吗？

根据您的描述，听起来您正在寻找指向指针的指针。

int **aofa;
aofa = malloc(sizeof(int*) * NUM_ARRAYS);
for (int i = 0 ; i != NUM_ARRAYS ; i++) {
    aofa[i] = malloc(sizeof(int) * getNumItemsInArray(i));
}
for (int i = 0 ; i != NUM_ARRAYS ; i++) {
    for (int j = 0 ; j != getNumItemsInArray(i) ; j++) {
        aofa[i][j] = i + j;
    }
}

NUM_ARRAYS数组可以具有不同数量的元素，如由getNumItemsInArray(i)函数返回的值所确定的。

int* ar[2];
int ar1[] = {1,2, 3};
int ar2[] = {5, 6, 7, 8, 9, 10};
ar[0] = ar1;
ar[1] = ar2;
cout << ar[1][2];

C版本，应该有助于澄清不同类型的声明：

#include <stdio.h>
int main()
{
/* let's make the arrays first                                             */
int array_A[3] = {1, 2, 3};
int array_B[3] = {4, 5, 6};
int array_C[3] = {7, 8, 9};
/* now let's declare some pointers to such arrays:                          */
int (*pA)[3] = &array_A;
int (*pB)[3] = &array_B;
int (*pC)[3] = &array_C;  /* notice the difference:                         */
/* int *pA[3] would be an array of 3 pointers to int because the [] operator*/
/* has a higher precedence than *(pointer) operator. so the statement would */
/* read: array_of_3 elements of type_pointer_to_int                         */
/* BUT, "int (*pA)[3]" is read: pointer_A (points to) type_array_of_3_ints! */
/* so now we need a different array to hold these pointers:                 */
/* this is called an_ARRAY_of_3_pointers to_type_array_of_3_ints            */
int (*ARRAY[3])[3] = {pA, pB, pC};
/* along with a a double pointer to type_array_of_3_ints:                   */
int (**PTR)[3] = ARRAY;
/* and check that PTR now points to the first element of ARRAY:             */
if (*PTR == pA) printf("PTR points to the first pointer from ARRAY n");
PTR++;
if (*PTR == pB) printf("PTR points to the second pointer from ARRAY! YAY!n");
   return 0;
}
> $ clang prog.c -Wall -Wextra -std=gnu89 "-ansi"  output:   
> PTR points to the first pointer from ARRAY 
> PTR points to the second pointer from ARRAY! YAY!

查看"指向对象数组的指针"一节http://www.functionx.com/cpp/Lesson24.htm它可能会对你有所帮助。

在C++中，您可以如下所示声明它。新操作符的工作方式与C中的malloc类似。

int** array = new int*[n];

#include <iostream>
using namespace std;
#define arraySize 3
const int arr1[] = {48,49,50};
const int arr2[] = {64,65,66};
const int arr3[] = {67,68,69};
typedef const int (*arrayByte);
arrayByte arrayPointer[arraySize] = {
    arr1,arr2,arr3
};
void printArr(const int arr[], int size){
    for(uint8_t x=0;x<size;x++){
        printf("value%d=%d n",x,arr[x]);
    }
}
int main()
{
    printf("Print Array 0n");
    printArr(arrayPointer[0],arraySize);
    printf("Print Array 1n");
    printArr(arrayPointer[1],arraySize);
    printf("Print Array 2n");
    printArr(arrayPointer[2],arraySize);
    return 0;
}

试试这个代码：C++在线

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