[英]Segmentation fault while dynamic allocation of 2D array in C
I am a newbie to C programming (relearning it after a long time) . 我是C编程的新手(经过很长时间的重新学习)。 I am trying to dynamically allocate memory to a 2D array using malloc. 我正在尝试使用malloc动态地将内存分配给2D数组。 I have tried following the answers on stackoverflow like this and this . 我曾尝试以下像计算器的答案这个和这个 。 But I still get the segmentation fault. 但我仍然得到分段错误。
My code is as below 我的代码如下
#include <stddef.h>
#include <stdio.h>
#include <stdlib.h>
void allocate2DArray(int **subset, int a, int b)
{
subset = (int **)malloc( a * sizeof(int *));
int i,j;
for(i = 0 ; i < a ; i++)
subset[i] = (int *) malloc( b * sizeof(int));
for(i = 0 ; i < a ; i++)
for(j = 0 ; j < b ; j++)
subset[i][j] = 0;
}
int main()
{
int **subset;
int a = 4, b = 4;
allocate2DArray(subset, a, b);
int i,j;
for( i = 0 ; i < a ; i++)
{
for( j = 0 ; j < b ; j++)
{
printf("%d ", subset[i][j]);
}
printf("\n");
}
}
When I comment the lines to print the array, it doens't give any error and program executes without segmentation fault. 当我注释行来打印数组时,它不会给出任何错误,程序执行时没有分段错误。 Please help me understand where I am going wrong. 请帮我理解我哪里错了。
All problems in computer science can be solved by another level of indirection : 计算机科学中的所有问题都可以通过另一层次的间接来解决 :
void allocate2DArray(int ***p, int a, int b)
{
int **subset;
*p = (int **) malloc(a * sizeof(int *));
subset = *p;
// ...
allocate2DArray(&subset, a, b);
you must pass a int ***subset
to the allocation function. 您必须将int ***subset
传递给分配函数。 This because arguments are passed by value. 这是因为参数是按值传递的。
You need this: 你需要这个:
void allocate2DArray(int ***subset, int a, int b)
and this: 和这个:
allocate2DArray(&subset, a, b);
By using int **subset;
通过使用int **subset;
it does not become 2D array. 它不会成为2D阵列。 It is still 1D storage and just pointer to pointer. 它仍然是1D存储,只是指向指针。
2D array means each element of a pointer buffer must point to a buffer which is suggested by ctn. 2D数组意味着指针缓冲区的每个元素必须指向由ctn建议的缓冲区。 He has suggested ***ptr and *ptr is malloced which created 1st dimension of buffer. 他建议*** ptr和* ptr是malloced,它创建了第一维缓冲区。 Now when you call allocate2DArray() again subset is allocated memory which create second dimension. 现在当你再次调用allocate2DArray()子集时,会分配内存来创建第二个维度。 Which validate my above statement - each element of pointer buffer must point to a buffer. 这验证了我的上述语句 - 指针缓冲区的每个元素都必须指向一个缓冲区。
so now with suggested code - 现在有了建议的代码 -
*p = (int **) malloc(a * sizeof(int *));
created an array each element of which point to buffer 'subset' which altogether create a true 2D array. 创建了一个数组,每个元素都指向缓冲区“子集”,它们共同创建了一个真正的2D数组。
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