数组指针的大小 [英] Sizeof Pointer to Array

问题描述

如果我有一个这样的数组：

  int a [3] [2] 
  

那么为什么是：

  sizeof（a + 0）== 8 
  

< >

  sizeof（a）== 24 
  

我不明白如何添加0到指针改变 sizeof 输出。

如果你添加 0 ，那么可能会有一些隐式类型转换？

a ，然后 a 首先转换为 int（*）类型的指针值[2] （指向 int [3] [2] 的数组的第一个元素。然后 0 被添加到它，它将 0 * sizeof（int [2]）那个指针的值。由于该乘法产生0，它将产生相同的指针值。因为它是一个指针， sizeof（a + 0）产生一个指针的大小，这是你的框上的8个字节。

如果你执行 sizeof（a），编译器没有理由转换 a 到一个指针值（只有当你想索引元素或做涉及元素地址的指针运算时才有意义）。因此，表达式 a 保持为数组类型，而改为 int [3] [2] 大小 int（*）[2] 。所以，你的框上的 3 * 2 * sizeof（int）是24字节。

希望这澄清事情。

If I have an array declared like this:

int a[3][2];

then why is:

sizeof(a+0) == 8

whereas:

sizeof(a)   == 24

I don't understand how adding 0 to the pointer changes the sizeof output. Is there maybe some implicit type cast?

解决方案

If you add 0 to a, then a is first converted to a pointer value of type int(*)[2] (pointing to the first element of an array of type int[3][2]). Then 0 is added to that, which adds 0 * sizeof(int[2]) bytes to the address represented by that pointer value. Since that multiplication yields 0, it will yield the same pointer value. Since it is a pointer, sizeof(a+0) yields the size of a pointer, which is 8 bytes on your box.

If you do sizeof(a), there is no reason for the compiler to convert a to a pointer value (that makes only sense if you want to index elements or to do pointer arithmetic involving the address of the elements). So expression a stays being of an array type, and you get the size of int[3][2] instead the size of int(*)[2]. So, 3 * 2 * sizeof(int) which on your box is 24 bytes.

Hope this clarifies things.

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