typedef 定长数组 [英] typedef fixed length array

问题描述

我必须定义一个 24 位数据类型.我使用 char[3] 来表示类型.我可以将 char[3] 定义为 type24 吗?我在代码示例中尝试过.我把 typedef char[3] type24; 放在我的头文件中.编译器没有抱怨它.但是当我在我的 C 文件中定义一个函数 void foo(type24 val) {} 时，它确实抱怨.我希望能够定义像 type24_to_int32(type24 val) 而不是 type24_to_int32(char value[3]) 之类的函数.

I have to define a 24-bit data type.I am using char[3] to represent the type. Can I typedef char[3] to type24? I tried it in a code sample. I put typedef char[3] type24; in my header file. The compiler did not complain about it. But when I defined a function void foo(type24 val) {} in my C file, it did complain. I would like to be able to define functions like type24_to_int32(type24 val) instead of type24_to_int32(char value[3]).

推荐答案

typedef 为

typedef char type24[3];

然而，这可能是一个非常糟糕的主意，因为结果类型是一个数组类型，但它的用户不会看到它是一个数组类型.如果用作函数参数，它将按引用传递，而不是按值传递，然后它的 sizeof 将是错误的.

However, this is probably a very bad idea, because the resulting type is an array type, but users of it won't see that it's an array type. If used as a function argument, it will be passed by reference, not by value, and the sizeof for it will then be wrong.

更好的解决方案是

typedef struct type24 { char x[3]; } type24;

您可能还想使用 unsigned char 而不是 char，因为后者具有实现定义的签名.

You probably also want to be using unsigned char instead of char, since the latter has implementation-defined signedness.

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