强制使用强类型检查在C（对于类型定义类型严格） [英] Enforce strong type checking in C (type strictness for typedefs)

问题描述

有没有办法强制执行相同类型的类型定义明确的转换？我已经处理UTF8，有时我感到困惑与指数的字符数和字节数。因此，它是很好，有一些类型定义：

Is there a way to enforce explicit cast for typedefs of the same type? I've to deal with utf8 and sometimes I get confused with the indices for the character count and the byte count. So it be nice to have some typedefs:

typedef unsigned int char_idx_t;
typedef unsigned int byte_idx_t;

在增加，你需要在它们之间进行明确的转换：

With the addition that you need an explicit cast between them:

char_idx_t a = 0;
byte_idx_t b;

b = a; // compile warning
b = (byte_idx_t) a; // ok

我知道，这样的功能不存在C，但也许你知道一个把戏或编译器扩展（preferable GCC）的做到这一点。

I know that such a feature doesn't exist in C, but maybe you know a trick or a compiler extension (preferable gcc) that does that.

修改
我还是真的不喜欢一般的匈牙利命名法。我不能使用它，因为项目编码约定的这个问题，但我现在用它在另一个类似的情形，在那里也是类型相同，含义非常相似。我不得不承认：它帮助。我永远不会去和一个开始宣布每个整数i，但在乔尔的榜样重叠类型，它可以挽救生命。

EDIT I still don't really like the Hungarian notation in general. I couldn't use it for this problem because of project coding conventions, but I used it now in another similar case, where also the types are the same and the meanings are very similar. And I have to admit: it helps. I never would go and declare every integer with a starting "i", but as in Joel's example for overlapping types, it can be life saving.

推荐答案

您可以这样做：

typedef struct {
    unsigned int c_idx;
} char_idx;

typedef struct {
    unsigned int b_idx;
} byte_idx;

然后，当你在使用每个你会看到：

Then you would see when you are using each:

char_idx a;
byte_idx b;

b.b_idx = a.c_idx;  

现在更清楚，它们是不同的类型，但仍然会进行编译。

Now it is more clear that they are different types but would still compile.

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