如何获得动态分配二维阵列的大小

问题描述

我已经动态分配的二维数组。这里是code

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);

现在我已经知道，有多少字节分配在arrofptr，arrofptr [0]，arrofptr [1]？有没有办法知道的大小？

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?

如果我们将打印

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

然后将打印4。

then it will print 4.

推荐答案

您无法找到 arrofptr 的大小，因为它只是一个指针的指针。您定义使用数组的数组。有没有办法，只有一个指针告诉大小的信息，你需要保持的尺寸的信息吧。

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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