使用NULL数组将内存分配给2D数组（c） [英] Allocating memory to 2D array using an array of NULL (c)
问题描述
感谢您抽出宝贵的时间阅读本文。
在我的问题中，向量定义为一维整数数组。
因此，向量数组应为
我被要求使用：
int **向量-2D数组
int size-一个整数，表示** vectors中有多少个向量
int * sizes-一个一维整数数组，表示向量的长度
thanks for taking the time in reading this.
In my question a "vector" is defined as a 1D dimensional array of integers.
Therefore an array of vectors would be a 2D dimensional array in which every vector can be of a different length.
I'm asked to use:
int** vectors- the 2D array
int size -an integer that represents how many vectors exist inside **vectors
int* sizes-a 1D array of integers that represents the length of the vectors
例如，对于：
向量= {{4,3,4,3}，{11,22,33,44,55,66}，NULL，{5}，{ 3,33,333,33,3}}。
的大小为5（向量内有5个向量）。
的大小为{4,6,0,1,5}（ 4是第一个向量的长度，依此类推。）
大小由用户在main（）的开头输入，并且** vectors和amp; * sizes是动态分配的，具有size的值。
for example,for:
vectors = {{4,3,4,3},{11,22,33,44,55,66},NULL,{5},{3,33,333,33,3}}.
size is 5 (there are 5 vectors inside vectors).
sizes is {4,6,0,1,5} (4 is the length of the first vector and so on).
size is inputted by the user at the beginning of main() and **vectors&*sizes are dynimacilly allocated with size's value.
我被要求编写函数：
int init（int *** vectors，int ** sizes，int size）初始化* * vectors为NULL数组，* vectors为零数组。
我想出了以下代码：
I'm asked to write the function:
int init(int ***vectors, int **sizes, int size) which initializes **vectors to be an array of NULLs and *sizes to be an array of zeros.
I came up with this code:
#define _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
#include <stdlib.h>
int init(int*** vectors, int** sizes, int size)
{
int i, k,j;
printf("check\n");
*vectors = (int**)malloc(size * sizeof(int*));
if (*vectors == NULL)
return 0;
for (i = 0; i < size; i++)
{
*(vectors + i) = NULL;
}
printf("check 2\n");
for (k = 0; k<size; k++)
{
if (*(vectors+k) != NULL)
printf("didn't work\n");
else
printf("current is null\n");
}
*sizes= (int*)malloc(size * sizeof(int));
if (*sizes == NULL)
return 0;
for (j= 0; j < size; j++)
{
*(sizes + j) = 0;
printf("%d ", *(sizes + j));
}
printf("\n");
return 1;
}
int main()
{
int size, i;
int** vectors = NULL;
int* sizes = NULL;
printf("\nPlease enter an amount of vectors:\n");
scanf("%d", &size);
printf("%d\n", init(&vectors, &sizes, size));
printf("size is %d now\n", size);
// for (i = 0; i < size; i++)
// printf("%d ", *(sizes+i));
printf("check 3\n");
free(sizes);
free(vectors);
printf("check 4\n");
printf("check 5\n");
return 0;
}
忘了说，如果init分配内存失败，它将返回0，否则返回1
打印支票是为了让我可以看到程序在哪里失败。
问题是无论如何，在打印最后一张支票（支票5）$ b $之后b程序失败。（运行时检查失败2）
如果有人可以帮助我了解我在做什么错，我将非常感激。
非常感谢您的阅读并度过了美好的一天
编辑：
i还在init内打印了数组大小/向量，只是看它是否打印零/空值，我实际上不需要这样做。
forgot to mention that init returns 0 if it fails to allocate memory and 1 otherwise.
printing the "checks" was so I could see where the program fails.
the problem is that no matter what,after printing the last check (check 5)
the program fails.(Run-Time Check Failure #2)
if anyone could help me understand what I'm doing wrong I would HIGHLY appreciate it.
thanks alot for reading and have an amazing day.
edit:
i also printed the array sizes/vectors inside init just to see if it prints zeros/nulls,i don't actually need to do it.
推荐答案
OP代码的一个问题是指针算法。给出：
One problem of OP's code is in the pointer arithmetic. Given:
int ***vectors;
*vectors = malloc(size * sizeof(int*));
此循环：
for (i = 0; i < size; i++)
{
*(vectors + i) = NULL;
}
将迭代下一个未分配的 pointer指针 strong>转换为int，而OP所需的是
Would iterate over the next unallocated pointer to pointer to pointer to int, while what the OP needs is
for (i = 0; i < size; i++)
{
*(*vectors + i) = NULL; // or (*vectors)[i] = NULL;
}
在以下循环中也是如此，其中 * （sizes + j）代替 *（* sizes + j）（或（* sizes）[j ] ）。
The same holds in the following loops, where *(sizes + j) is used instead of *(*sizes + j) (or (*sizes)[j]).
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