在函数中分配多维数组 [英] malloc-ating multidimensional array in function

问题描述

我正在尝试在 C 程序中分配一个二维数组.它在这样的主要功能中工作正常(如 这里):

I'm trying to allocate a 2d array in a C program. It works fine in the main function like this (as explained here):

#include <stdio.h>
#include <stdlib.h>

int main(int argc, char ** argv)
{
    int ** grid;
    int i, nrows=10, ncols=10;
    grid = malloc( sizeof(int *) * nrows);

    if (grid == NULL){
        printf("ERROR: out of memory
");
        return 1;
    }

    for (i=0;i<nrows;i++){
        grid[i] = malloc( sizeof(int) * ncols);
        if (grid[i] == NULL){
            printf("ERROR: out of memory
");
            return 1;
        }
    }
    printf("Allocated!
");

    grid[5][6] = 15;
    printf("%d
", grid[5][6]);
    return 0;
}

但是由于我必须使用不同的数组多次执行此操作，因此我试图将代码移至单独的函数中.

But since I have to do this several times with different arrays, I was trying to move the code into a separate function.

#include <stdio.h>
#include <stdlib.h>

int malloc2d(int ** grid, int nrows, int ncols){
    int i;
    grid = malloc( sizeof(int *) * nrows);

    if (grid == NULL){
        printf("ERROR: out of memory
");
        return 1;
    }

    for (i=0;i<nrows;i++){
        grid[i] = malloc( sizeof(int) * ncols);
        if (grid[i] == NULL){
            printf("ERROR: out of memory
");
            return 1;
        }
    }
    printf("Allocated!
");
    return 0;
}

int main(int argc, char ** argv)
{
    int ** grid;

    malloc2d(grid, 10, 10);
    grid[5][6] = 15;
    printf("%d
", grid[5][6]);
    return 0;
}

然而，虽然它在分配时没有抱怨，但在访问数组时出现分段错误.我阅读了关于衰变数组和类似主题的不同帖子，但我仍然无法弄清楚如何解决这个问题.我想我没有正确地将二维数组传递给函数.

However, although it doesn't complain while allocating, I get segmentation fault when accessing the array. I read different posts on decayed arrays and similar topics, but I still can't figure out how to solve this problem. I imagine I'm not passing the 2d array correctly to the function.

非常感谢.

推荐答案

那不是多维数组；它是一个一维数组，包含指向一维数组的指针.多维数组不包含指针；它们是单个内存块.

That is not a multidimensional array; it is a single dimensional array containing pointers to single dimensional arrays. Multidimensional arrays do not contain pointers; they are single memory blocks.

您的问题是您有一个指向指针的指针，并且您正试图通过参数从您的函数中返回它.如果你打算这样做，你将需要一个指向一个指针的指针作为你的参数，你将不得不将一个指针的地址传递给一个指向该方法的指针.如果不这样做，则不会更改 main 中变量 grid 的值——而是将作为参数复制的值更改为malloc2d 函数.由于 main 中的 grid 未初始化，您会得到未定义的行为.

Your problem here is that you have a pointer to a pointer, and you're trying to return it from your function through a parameter. If you're going to do that, you're going to need a pointer to a pointer to a pointer as your parameter, and you're going to have to pass the address of a pointer to a pointer to the method. If you don't do this, you're not changing the value of the variable grid in main -- you're changing the one which was copied as the parameter to the malloc2d function. Because the grid in main is left uninitialized, you get a undefined behavior.

以下是我所说的修复示例:

Here's an example of what I mean as the fix:

#include <stdio.h>
#include <stdlib.h>

int malloc2d(int *** grid, int nrows, int ncols){
    int i;
    *grid = malloc( sizeof(int *) * nrows);

    if (*grid == NULL){
        printf("ERROR: out of memory
");
        return 1;
    }

    for (i=0;i<nrows;i++){
        (*grid)[i] = malloc( sizeof(int) * ncols);
        if ((*grid)[i] == NULL){
            printf("ERROR: out of memory
");
            return 1;
        }
    }
    printf("Allocated!
");
    return 0;
}

int main(int argc, char ** argv)
{
    int ** grid;

    malloc2d(&grid, 10, 10);
    grid[5][6] = 15;
    printf("%d
", grid[5][6]);
    return 0;
}

附加说明:

• 如果单个分配失败，则会泄漏第一个数组的分配以及所有先前行的分配.您需要在返回之前调用 free.
• 您通过一个参数返回，即使您真的没有必要这样做.如果我在写这个，我会让方法返回 int **，并通过返回 0 来表示错误.

• If a single allocation fails, you leak the allocation for the first array, as well as the allocations for all previous rows. You need to call free on those before returning.
• You're returning through a parameter, even though you really don't have to. If I were writing this, I'd make the method return int **, and signal error by returning 0.

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