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。有任何想法吗?
答案 0 :(得分:4)
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];
答案 1 :(得分:2)
从你的描述来看,听起来你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
数组可能具有不同数量的元素,由getNumItemsInArray(i)
函数返回的值确定。
答案 2 :(得分:0)
查看“指向对象阵列的指针”部分 http://www.functionx.com/cpp/Lesson24.htm 它可能会对你有帮助。
答案 3 :(得分:0)
在C ++中,您可以如下所示进行声明。new运算符的工作方式类似于C中的malloc。
int** array = new int*[n];
答案 4 :(得分:0)
C版本,应该有助于阐明不同类型的声明:
#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!