为什么2维数组传递给函数需要它的大小？ [英] Why does 2 dimension array passing to function requires its size?

问题描述

启发<一个href=\"http://stackoverflow.com/questions/9446707/correct-way-of-passing-2-dimensional-array-into-a-function\">this quiestion。

Inspired this quiestion .

为什么 -

void display(int p[][SIZE]) // allowed

和

void display(int p[][]) // not allowed

推荐答案

由于阵列衰减的的指针。如果你不提供阵列的第二个维度的基数，编译器不知道如何取消引用这个指针。

Because arrays decay to pointers when passed to a function. If you do not provide the cardinality of the second dimension of the array, the compiler would not know how to dereference this pointer.

下面是一个较长的解释是：当你写这个

Here is a longer explanation: when you write this

p[index]

编译器执行一些指针运算找到的元素，它需要引用地址：它乘以首页按大小 P 的元素，并将其添加到基址 p ：

the compiler performs some pointer arithmetic to find the address of the element that it needs to reference: it multiplies index by the size of p's element, and adds it to the base address of p:

address = <base address of p> + index * <size of p's element>

当你试图传递一个数组这样， P [] [] ，编译器只知道 P ，而不是它的元件的尺寸。这是为了让编译器知道 P 的元素，你需要提供第二个维度的基数大小。

When you try passing an array like this, p[][], the compiler knows only the base address of p, but not the size of its element. It is in order for the compiler to know the size of p's element that you need to provide the cardinality of the second dimension.

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