分配的二维数组用C [英] Allocating 2-D array in C

问题描述

我想在运行时分配在C 2-D数组。现在此所用的常规方式这样来实现：

I want to allocate a 2-D array in C at runtime. Now this can be achieved in the conventional manner like this:

int *matrix[rows]
for (row = 0; row < rows; ++row) {
  matrix[row] = (int *)malloc(ncol*sizeof(int));
}

但我发现了另一个方法，做同样的事情：

But I found another method, which does the same thing:

int (*p)[rows];
p=(int (*)[rows])malloc(rows*cols*sizeof(int));

谁能解释第二个声明是如何工作的？具体来说，什么是的意思（INT（*）[行]）的malloc ？据我所知，的malloc 用于像为（int *）malloc的（NCOL *的sizeof（INT））或（字符*）malloc的（NCOL *的sizeof（char）的）。

Can anyone explain how the 2nd declaration works? Specifically, what is meant by (int (*)[rows])malloc? To the best of my knowledge, malloc is used like (int *)malloc(ncol*sizeof(int)) or (char *)malloc(ncol*sizeof(char)).

推荐答案

在这里，你投的malloc 的返回值的类型的指针数组行 INT > 。

Here, you cast malloc's return value to the type pointer to array rows of int.

顺便说一句，在C，指针的强制转换为无效的指针对象没有该电源线，甚至是无用的。你不应该担心这些细节。下面code工作确实也是如此。

By the way, in C, the cast of a pointer to void to a pointer to object is not requiered, and even useless. You should not worry about these details. The following code works indeed as well.

#include <stdlib.h>

int (*p)[rows];
p = malloc(rows * cols * sizeof(int));

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