和了``＆放大器之间的区别`;在了`C / C ++其中`了`是一个数组 [英] Difference between `a` and `&a` in C/C++ where `a` is an array

问题描述

我感到困惑以下code的输出。

I am confused about the output of the following code.

#include<iostream>
#include<cstdlib>
using namespace std;
int main()
{
  int a[] = {1,2,3};
  cout << a         << "  " << &a         << endl;
  cout << sizeof(a) << "  " << sizeof(&a) << endl;
  return 0;
}

的输出是

0xbfcd3ae4  0xbfcd3ae4
12  4

哪有 A 和＆放大器;一个打印相同的前pression，但有不同的大小？
我始终认为，任何阵列，它的名字总是有第一个字节的值=地址。

How can a and &a print the same expression but have different sizes? I always thought that for any array, its name always has the value = address of the first byte.

还有＆放大器;一个应该没有什么意义，因为一个不能有一个地址（用＆放大器获得;运营商）到地址（的价值）。然而，code给出了一个输出和逸岸'一==＆安培;一个'
根据输出。

Also &a should not make sense, since one cannot have an address (obtained with the & operator) to an address(the value of a). Yet the code gives an output and infact 'a == &a' according to the output.

同样为什么的输出的sizeof（A）= 12 （这是总的内存占用）
由阵列？ A 是一个指针本身的sizeof（A）= 4字节（在我的32位的Ubuntu 11.04）

Similarly why is the output of sizeof(a) = 12 (which is the total memory occupied) by the array? a being a "pointer" itself sizeof(a) = 4 bytes (on my 32 bit Ubuntu 11.04)

显然是有我有一些误解。莫非有人排序了这一点给我吗？

Obviously there is some misconception I am having. Could some one sort this out for me ?

推荐答案

这是数组不是指针，而是一个数组的衰减到的当您尝试使用它像一个指针。你的情况自动打印数组的地址，将其转换成一个指针。

An array is not a pointer, but an array decays to a pointer when you try to use it like one. In your case printing the address of the array automatically converts it into a pointer.

有之间的自动转换指针并与＆放大器明确创建所述一个差别不大; ，除了一个是指向单个元件，而另一个是一个指向整个阵列。如果你已经使用＆放大器;一个[0] 那么他们将是相同的。

There's little difference between the automatically converted pointer and the one created explicitly with &, except that one is a pointer to a single element while the other is a pointer to the entire array. If you had used &a[0] then they would be identical.

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