数组的地址 [英] Address of array

问题描述

int t[10];

int * u = t;

cout << t << " " << &t << endl;

cout << u << " " << &u << endl;

输出：

0045FB88 0045FB88
0045FB88 0045FB7C

为 U 输出是有道理的。

据我所知， T 和＆amp; T公司[0] 应具有相同的价值，但怎么来的＆amp; T公司也是一样的吗？ ＆是什么放大器;？吨实际上意味着

I understand that t and &t[0] should have the same value, but how come &t is also the same? What does &t actually mean?

推荐答案

在 T 在EX pression用于自身，一个数组用于─指针转换发生，这将产生一个指针数组的第一元素

When t is used on its own in the expression, an array-to-pointer conversion takes place, this produces a pointer to the first element of the array.

在 T 被用作＆放大器的参数; 运营商，没有这样的转换发生。在＆安培; 则明确采取 T 的地址（阵列）。 ＆放大器;：T 是一个指向数组作为一个整体

When t is used as the argument of the & operator, no such conversion takes place. The & then explicitly takes the address of t (the array). &t is a pointer to the array as a whole.

阵列的第一个元素是在存储器中的相同位置的整个阵列的开始，所以这两个指针的值相同。

The first element of the array is at the same position in memory as the start of the whole array, and so these two pointers have the same value.

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