char 和 char[1] 的区别
在 C++ 中，使用 char 和 char[1] 有什么区别(如果有的话).
In C++ what is the difference (if any) between using char and char[1].
例子:
struct SomeStruct
{
char x;
char y[1];
};
对于 unsigned char 是否有同样的原因?
Do the same reasons follow for unsigned char?
推荐答案
主要区别只是用于访问单个字符的语法.
The main difference is just the syntax you use to access your one char.
我所说的访问"是指使用语言中的各种运算符对其进行操作，与 char数组.这听起来好像 x 和 y 几乎完全不同.如果事实上它们都由"一个字符组成，但该字符以非常不同的方式表示.
By "access" I mean, act upon it using the various operators in the language, most or all of which do different things when applied to a char compared with a char array. This makes it sound as if x and y are almost entirely different. If fact they both "consist of" one char, but that char has been represented in a very different way.
实现可能会导致其他差异，例如它可能会根据您使用的结构以不同的方式对齐和填充结构.但我怀疑它会.
The implementation could cause there to be other differences, for example it could align and pad the structure differently according to which one you use. But I doubt it will.
运算符差异的一个例子是 char 是可赋值的，而数组不是:
An example of the operator differences is that a char is assignable, and an array isn't:
SomeStruct a;
a.x = 'a';
a.y[0] = 'a';
SomeStruct b;
b.x = a.x; // OK
b.y = a.y; // not OK
b.y[0] = a.y[0]; // OK
但 y 不可赋值这一事实并不能阻止 SomeStruct 可赋值:
But the fact that y isn't assignable doesn't stop SomeStruct being assignable:
b = a; // OK
所有这些都与类型无关，char 与否.一个类型的对象和一个大小为 1 的类型的数组，就内存中的内容而言几乎相同.
All this is regardless of the type, char or not. An object of a type, and an array of that type with size 1, are pretty much the same in terms of what's in memory.
顺便说一句，您在 char 和 char[1] 中使用"它会产生很大的不同，有时会有所帮助使人们误以为数组实际上是指针.不是您的示例，而是作为函数参数:
As an aside, there is a context in which it makes a big difference which you "use" out of char and char[1], and which sometimes helps confuse people into thinking that arrays are really pointers. Not your example, but as a function parameter:
void foo(char c); // a function which takes a char as a parameter
void bar(char c[1]); // a function which takes a char* as a parameter
void baz(char c[12]); // also a function which takes a char* as a parameter
bar 和 baz 的声明中提供的数字被 C++ 语言完全忽略.显然有人在某些时候觉得它作为一种文档形式对程序员很有用，表明函数 baz 期望它的指针参数指向 12 字符数组的第一个元素.
The numbers provided in the declarations of bar and baz are completely ignored by the C++ language. Apparently someone at some point felt that it would be useful to programmers as a form of documentation, indicating that the function baz is expecting its pointer argument to point to the first element of an array of 12 char.
在 bar 和 baz 中，c 从来没有数组类型――它看起来像一个数组类型，但实际上不是，它只是一种花哨的特殊情况语法，与 含义相同字符 *c.这就是为什么我将引号放在使用"上 - 你根本没有真正使用 char[1]，它只是看起来像它.
In bar and baz, c never has array type - it looks like an array type, but it isn't, it's just a fancy special-case syntax with the same meaning as char *c. Which is why I put the quotation marks on "use" - you aren't really using char[1] at all, it just looks like it.
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