如何获取动态分配的二维数组的大小 [英] How to get the size of dynamically allocated 2d array

问题描述

我已经动态分配了二维数组.这是代码

I have dynamically allocated 2D array. Here is the code

int **arrofptr ;
arrofptr = (int **)malloc(sizeof(int *) * 2);
arrofptr[0] = (int *)malloc(sizeof(int)*6144);
arrofptr[1] = (int *)malloc(sizeof(int)*4800);

现在我必须知道在 arrofptr,arrofptr[0],arrofptr[1] 中分配了多少字节?有没有办法知道尺寸?

Now i have to know that how many bytes are allocated in arrofptr,arrofptr[0],arrofptr[1]? is there any way to know the size?

如果我们要打印

sizeof(arrofptr);
sizeof(arrofptr[0]);
sizeof(arrofptr[1]);

然后它会打印 4.

推荐答案

你找不到arrofptr 的大小，因为它只是一个指向指针的指针.您正在使用它定义一个数组数组.没有办法只用一个指针来告诉size信息，需要自己维护size信息.

You can't find size of arrofptr, because it is only a pointer to pointer. You are defining an array of arrays using that. There's no way to tell the size information with only a pointer, you need to maintain the size information yourself.

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