实现一个字符串的编译时间检查机制的独特性 [英] Implementing compile-time mechanism checking uniqueness of a string

问题描述

定义我的问题的最简单的方法是，我试图实现一种机制，将检查是否相同的字符串已经使用（或一对（数字，字符串））。我想用C preprocessor一个聪明的方式来实现这种机制。我也想，这个机制给了我编译错误时有冲突或运行时在调试模式下的错误（通过检查断言）。我们不希望开发者加入消息时，因为每个消息应该是唯一犯了一个错误。我知道，它可以通过计算散列或例如完成的 CRC / MD5 但是这个机制将是冲突的脆弱，我需要避免的。至关重要的是，所有的消息只能使用一次。

The simplest way of defining my problem is that I'm trying to implement a mechanism that would check whether the same string had already been used (or a pair (number, string)). I would like this mechanism to be implemented in a smart way using C preprocessor. I would also like that this mechanism gave me compile errors when there is a conflict or run-time errors in Debug mode (by checking assertions). We don't want the developer to make a mistake when adding a message, as every message should be unique. I know that it could be done by calculating a hash or for example crc/md5 but this mechanism would be conflict-vulnerable which I need to avoid. It is crucial that every message can be used only once.

这种机制的例子行为：

addMessage(1, "Message1") //OK 
addMessage(2, "Message2") //OK 
. 
. 
. 
addMessage(N, "MessageN") //OK 
addMessage(2, "Message2") //Compile error, Message2 has already been used 

另类的行为（在调试的时候code）：

Alternative behaviour (when Debugging code):

addMessage(1, "Message1") //OK 
addMessage(2, "Message2") //OK 
. 
. 
. 
addMessage(N, "MessageN") //OK 
addMessage(2, "Message2") //Assertion failed, because Message2 has already been used 

这样做会的#define 和和#undef 指令的智能使用的preferred方式。在一般的preprocessor应以一个巧妙的方法（我不知道这是可能的），也许它可以通过宏的适当组合可以实现使用？任何C preprocessor黑客，可以帮助我解决这个问题？

The preferred way of doing it would be smart usage of #define and #undef directives. In general the preprocessor should be used in a smart way (I am not sure if this is possible) maybe it can be achieved by appropriate combinations of macros? Any C preprocessor hacker that could help me solve this problem?

//编辑：我需要这些信息是唯一的全球性，不仅是code块内（如同if语句功能）。

// I need those messages to be unique globally, not only inside one code block (like function of if-statement).

// EDIT2：这个问题的最好描述是，我有100个不同的源文件，我想检查了preprocessor（或者不是在开始分析与脚本源文件等其他可能的机制的每一次编辑，这将是非常耗时且会增加另一级到一个足够复杂项目）如果一个字符串（或一个preprocessor定义中），使用一个以上的时间。我仍然不知道如何做到这一点（我知道这可能不是在所有可行，但我希望它实际上是）。

// The best description of the problem would be that I have 100 different source files and I would like to check with a preprocessor (or possibly other mechanism other than parsing source files with a script at a start of the compilation every-time, which would be very time-consuming and would add another stage to an enough complicated project) if a string (or a preprocessor definition) was used more than one time. I still have no idea how to do it (I know it may not be possible at all but I hope it actually is).

推荐答案

这会给出错误的重复的字符串：

This will give an error on duplicate strings:

constexpr bool isequal(char const *one, char const *two) {
  return (*one && *two) ? (*one == *two && isequal(one + 1, two + 1))
    : (!*one && !*two);
}

constexpr bool isunique(const char *test, const char* const* list)
{
    return *list == 0 || !isequal(test, *list) && isunique(test, list + 1);
}

constexpr int no_duplicates(const char* const* list, int idx)
{
    return *list == 0 ? -1 : (isunique(*list, list + 1) ? no_duplicates(list + 1, idx + 1) : idx);
}

template <int V1, int V2> struct assert_equality
{
    static const char not_equal_warning = V1 + V2 + 1000;
};

template <int V> struct assert_equality<V, V>
{
    static const bool not_equal_warning = 0;
};

constexpr const char* l[] = {"aa", "bb", "aa", 0};
static_assert(assert_equality<no_duplicates(l, 0), -1>::not_equal_warning == 0, "duplicates found");

这是G ++输出：

g++ -std=c++11 unique.cpp 
unique.cpp: In instantiation of ‘const char assert_equality<0, -1>::not_equal_warning’:
unique.cpp:29:57:   required from here
unique.cpp:20:53: warning: overflow in implicit constant conversion [-Woverflow]
unique.cpp:29:1: error: static assertion failed: duplicates found

第一个模板参数（在这种情况下，0），以assert_equality告诉你一个重复的字符串的拳头地位。

The first template parameter (in this case 0) to 'assert_equality' tells you the fist position of a duplicate string.

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