数组地址 [英] Address of an 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

u 的输出是有意义的.

我知道 t 和 &t[0] 应该具有相同的值，但是为什么 &t 也是相同的?&t 实际上是什么意思?

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

当 t 在表达式中单独使用时，会发生数组到指针的转换，这会产生一个指向数组的第一个元素.

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.

当 t 用作 & 运算符的参数时，不会发生此类转换.& 然后显式地获取 t(数组)的地址.&t 是一个指向整个数组的指针.

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.

数组的第一个元素与整个数组的开头在内存中的相同位置，因此这两个指针具有相同的值.

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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