为什么sizeof（array）在这两种情况下不同？ [英] Why is sizeof(array) different in these two cases?

问题描述

可能重复：

Sizeof array passed as parameter

我目前的系统（似乎是64位给定的指针大小）打印：

Given the following code, the system I'm currently on (which seems to be 64-bit given the size of a pointer) prints:

32
4

现在，此系统上的int大小应为4字节，因此8 * 4 = 32有意义。另外，我理解指针的大小也应该是4个字节，这是我猜测是发生在foo中的数组。

Now, the size of an int should be 4 bytes on this system, so 8*4 = 32 makes sense. Also, I understand that the size of a pointer should also be 4 bytes, which is what I'm guessing is happening to the array in foo.

我的问题是为什么是sizeof（arr）在foo函数中处理不同于sizeof（myarray）在main（）当两个都是专门int [8]数组？我预计在这两种情况下都会收到32，它会让我为什么会以其他方式结束。

My question is why is sizeof(arr) in the foo function treated differently than sizeof(myarray) in main() when both are specifically int[8] arrays? I had anticipated receiving 32 in both cases and it confuses me why it would end up otherwise.

#include <iostream>

void foo(int arr[8])
{
    std::cerr << sizeof(arr) << std::endl;
}

int main()
{
    int myarray[8];
    std::cerr << sizeof(myarray) << std::endl
    foo(myarray);
}

推荐答案

数组在传递时衰减成指针到函数中，以便在 foo（）中打印4个字节的int数组指针 sizeof 。

Arrays decay into pointers when passed into functions such that you are printing the sizeof the array pointer of int of 4 bytes in foo().

sizeof（myarray）输出数组的大小为32（num元素x int大小） 。

The sizeof(myarray) prints the size of the array which is 32 (num elements x size of int).

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