问题描述
如果您使用 C，我相信我理解普通变量和指针在内存中的表示方式.
I believe I understand how normal variables and pointers are represented in memory if you are using C.
例如，很容易理解一个指针Ptr会有一个地址，它的值会是一个不同的地址，也就是它指向的内存空间.代码如下:
For example, it's easy to understand that a pointer Ptr will have an address, and its value will be a different address, which is the space in memory it's pointing to. The following code:
int main(){
int x = 10;
int *Ptr;
Ptr = &x;
return 0;
}
在内存中会有以下表示:
Would have the following representation in memory:
+---------------------+-------------+---------+
| Variable Name | Address | Value |
+---------------------+-------------+---------+
| x | 3342 | 10 |
+---------------------+-------------+---------+
| Ptr | 5466 | 3342 |
+---------------------+-------------+---------+
然而，我发现很难理解数组在内存中的表示方式.例如代码:
However I find it difficult to understand how arrays are represented in memory. For example the code:
int main(){
int x[5];
x[0]=12;
x[1]=13;
x[2]=14;
printf("%p\n",(void*)x);
printf("%p\n",(void*)&x);
return 0;
}
两次输出相同的地址(为简单起见，10568).这意味着 x==&x.然而*x(或数组符号中的x[0])等于12，*(x+1)(或数组符号中的x[1])等于13，依此类推.这如何表示?一种方法可能是这样的:
outputs the same address twice (for the sake of simplicity 10568). Meaning that x==&x. Yet *x (or x[0] in array notation) is equal to 12, *(x+1) (or x[1] in array notation) is equal to 13 and so on. How can this be represented? One way could be this:
+---------------------+-------------+----------+----------------------+
| Variable Name | Address | Value | Value IF array |
+---------------------+-------------+----------+----------------------+
| x | 10568 | 10568 | 12 |
+---------------------+-------------+----------+----------------------+
| | 10572 | | 13 |
+---------------------+-------------+----------+----------------------+
| | 10576 | | 14 |
+---------------------+-------------+----------+----------------------+
| | 10580 | | trash |
+---------------------+-------------+----------+----------------------+
| | 10584 | | trash |
+---------------------+-------------+----------+----------------------+
这是接近发生的事情，还是完全没有?
Is this close to what happens, or completely off?
推荐答案
数组是一组连续的对象，中间没有空格.这意味着第二个示例中的 x 在内存中表示为:
An array is a block of contiguous objects with no spaces in between. This means that x in your second example is represented in memory as:
+---------------------+-------------+---------+
| Variable Name | Address | Value |
+---------------------+-------------+---------+
| x | 10568 | 12 |
| | +---------+
| | | 13 |
| | +---------+
| | | 14 |
| | +---------+
| | | ?? |
| | +---------+
| | | ?? |
+---------------------+-------------+---------+
也就是说，x 有五个 int 大，并且只有一个地址.
That is, x is five ints big, and has a single address.
关于数组的奇怪部分不在于它们的存储方式——而是它们在表达式中的计算方式.如果您在某处使用不是一元 & 或 sizeof 运算符的主题的数组名称，则其计算结果为其第一个成员的地址.
The weird part about arrays isn't in how they're stored - it's how they're evaluated in expressions. If you use an array name somewhere that it isn't the subject of the unary & or sizeof operators, it evaluates to the address of its first member.
也就是说，如果你只写x，你会得到一个类型为int *的值10568.
That is, if you just write x, you will get a value 10568 with type int *.
另一方面，如果您编写 &x，则特殊规则不适用 - 因此 & 运算符的工作方式与通常一样，即意味着它获取数组的地址.在示例中，这将是一个类型为 int (*)[5] 的值 10568.
If, on the other hand you write &x, then the special rule doesn't apply - so the & operator works like it normally does, which means that it fetches the address of the array. In the example, this will be a value 10568 with type int (*)[5].
x == &x 的原因是数组第一个成员的地址必须等于数组本身的地址，因为数组从它的第一个成员开始.
The reason that x == &x is that the address of the first member of an array is necessarily equal to the address of the array itself, since an array starts with its first member.
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