可变递归预处理器宏 - 可能吗？

发布于 2024-07-19 08:14:40 字数 752 浏览 5 评论 0原文

我遇到了一些理论问题。在我维护的一段代码中，有一组宏，例如

#define MAX_OF_2(a, b)       (a) > (b) ? (a) : (b)
#define MAX_OF_3(a, b, c)    MAX_OF_2(MAX_OF_2(a, b), c)
#define MAX_OF_4(a, b, c, d) MAX_OF_2(MAX_OF_3(a, b, c), d)
...etc up to MAX_OF_8

我想做的是将它们替换为如下内容：

/* Base case #1, single input */
#define MAX_OF_N(x)      (x)

/* Base case #2, two inputs */
#define MAX_OF_N(x, y)   (x) > (y) ? (x) : (y)

/* Recursive definition, arbitrary number of inputs */
#define MAX_OF_N(x, ...) MAX_OF_N(x, MAX_OF_N(__VA_ARGS__))

...当然，这不是有效的预处理器代码。

忽略这种特殊情况可能应该使用函数而不是预处理器宏来解决，是否可以定义可变参数 MAX_OF_N() 宏？

为了清楚起见，最终结果应该是一个单独的宏，它采用任意数量的参数并计算其中最大的参数。我有一种奇怪的感觉，这应该是可能的，但我不知道如何实现。

原文

I've run into a little theoretical problem. In a piece of code I'm maintaining there's a set of macros like

#define MAX_OF_2(a, b)       (a) > (b) ? (a) : (b)
#define MAX_OF_3(a, b, c)    MAX_OF_2(MAX_OF_2(a, b), c)
#define MAX_OF_4(a, b, c, d) MAX_OF_2(MAX_OF_3(a, b, c), d)
...etc up to MAX_OF_8

What I'd like to do is replace them with something like this:

/* Base case #1, single input */
#define MAX_OF_N(x)      (x)

/* Base case #2, two inputs */
#define MAX_OF_N(x, y)   (x) > (y) ? (x) : (y)

/* Recursive definition, arbitrary number of inputs */
#define MAX_OF_N(x, ...) MAX_OF_N(x, MAX_OF_N(__VA_ARGS__))

...which, of course, is not valid preprocessor code.

Ignoring that this particular case should probably be solved using a function rather than a preprocessor macro, is it possible to define a variadic MAX_OF_N() macro?

Just for clarity, the end result should be a single macro that takes an arbitrary number of parameters and evaluates to the largest of them. I've got an odd feeling that this should be possible, but I'm not seeing how.

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悲凉≈ 2024-07-26 08:14:40

可以编写一个计算其调用参数数量的宏。（我找不到第一次看到它的链接。）所以你可以编写 MAX_OF_N() ，它可以按照你的意愿工作，但你仍然需要所有编号的宏，直到达到一定的限制：

#define MAX_OF_1(a)         (a)         
#define MAX_OF_2(a,b)       max(a, b)

#define MAX_OF_3(a,...)    MAX_OF_2(a,MAX_OF_2(__VA_ARGS__))
#define MAX_OF_4(a,...)    MAX_OF_2(a,MAX_OF_3(__VA_ARGS__))
#define MAX_OF_5(a,...)    MAX_OF_2(a,MAX_OF_4(__VA_ARGS__))
...
#define MAX_OF_64(a,...)   MAX_OF_2(a,MAX_OF_63(__VA_ARGS__))

// NUM_ARGS(...) evaluates to the literal number of the passed-in arguments.
#define _NUM_ARGS2(X,X64,X63,X62,X61,X60,X59,X58,X57,X56,X55,X54,X53,X52,X51,X50,X49,X48,X47,X46,X45,X44,X43,X42,X41,X40,X39,X38,X37,X36,X35,X34,X33,X32,X31,X30,X29,X28,X27,X26,X25,X24,X23,X22,X21,X20,X19,X18,X17,X16,X15,X14,X13,X12,X11,X10,X9,X8,X7,X6,X5,X4,X3,X2,X1,N,...) N
#define NUM_ARGS(...) _NUM_ARGS2(0, __VA_ARGS__ ,64,63,62,61,60,59,58,57,56,55,54,53,52,51,50,49,48,47,46,45,44,43,42,41,40,39,38,37,36,35,34,33,32,31,30,29,28,27,26,25,24,23,22,21,20,19,18,17,16,15,14,13,12,11,10,9,8,7,6,5,4,3,2,1,0)

#define _MAX_OF_N3(N, ...) MAX_OF_ ## N(__VA_ARGS__)
#define _MAX_OF_N2(N, ...) _MAX_OF_N3(N, __VA_ARGS__)
#define MAX_OF_N(...)      _MAX_OF_N2(NUM_ARGS(__VA_ARGS__), __VA_ARGS__)

现在MAX_OF_N(a,b,c,d,e) 将计算为 max(a, max(b, max(c, max(d, e)))) 。（我已经在 gcc 4.2.1 上进行了测试。）

请注意，基本情况 (MAX_OF_2) 在扩展中不要多次重复其参数（这就是为什么我将 在本例中为 >max）。否则，您将在每个级别将扩展的长度加倍，因此您可以想象 64 个参数会发生什么:)

It's possible to write a macro that evaluates to the number of arguments it's called with. (I couldn't find a link to the place where I first saw it.) So you could write MAX_OF_N() that would work as you'd like, but you'd still need all the numbered macros up until some limit:

#define MAX_OF_1(a)         (a)         
#define MAX_OF_2(a,b)       max(a, b)

#define MAX_OF_3(a,...)    MAX_OF_2(a,MAX_OF_2(__VA_ARGS__))
#define MAX_OF_4(a,...)    MAX_OF_2(a,MAX_OF_3(__VA_ARGS__))
#define MAX_OF_5(a,...)    MAX_OF_2(a,MAX_OF_4(__VA_ARGS__))
...
#define MAX_OF_64(a,...)   MAX_OF_2(a,MAX_OF_63(__VA_ARGS__))

// NUM_ARGS(...) evaluates to the literal number of the passed-in arguments.
#define _NUM_ARGS2(X,X64,X63,X62,X61,X60,X59,X58,X57,X56,X55,X54,X53,X52,X51,X50,X49,X48,X47,X46,X45,X44,X43,X42,X41,X40,X39,X38,X37,X36,X35,X34,X33,X32,X31,X30,X29,X28,X27,X26,X25,X24,X23,X22,X21,X20,X19,X18,X17,X16,X15,X14,X13,X12,X11,X10,X9,X8,X7,X6,X5,X4,X3,X2,X1,N,...) N
#define NUM_ARGS(...) _NUM_ARGS2(0, __VA_ARGS__ ,64,63,62,61,60,59,58,57,56,55,54,53,52,51,50,49,48,47,46,45,44,43,42,41,40,39,38,37,36,35,34,33,32,31,30,29,28,27,26,25,24,23,22,21,20,19,18,17,16,15,14,13,12,11,10,9,8,7,6,5,4,3,2,1,0)

#define _MAX_OF_N3(N, ...) MAX_OF_ ## N(__VA_ARGS__)
#define _MAX_OF_N2(N, ...) _MAX_OF_N3(N, __VA_ARGS__)
#define MAX_OF_N(...)      _MAX_OF_N2(NUM_ARGS(__VA_ARGS__), __VA_ARGS__)

Now MAX_OF_N(a,b,c,d,e) will evaluate to max(a, max(b, max(c, max(d, e)))). (I've tested on gcc 4.2.1.)

Note that it's critical that the base case (MAX_OF_2) doesn't repeat its arguments more than once in the expansion (which is why I put max in this example). Otherwise, you'd be doubling the length of the expansion for every level, so you can imagine what will happen with 64 arguments :)

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蓝咒 2024-07-26 08:14:40

您可能会认为这是作弊，因为它不是递归的，并且它不在预处理器中完成工作。它使用 GCC 扩展。而且它只适用于一种类型。然而，它是一个可变参数 MAX_OF_N 宏：

#include <iostream>
#include <algorithm>

#define MAX_OF_N(...) ({\
        int ra[] = { __VA_ARGS__ }; \
        *std::max_element(&ra[0], &ra[sizeof(ra) / sizeof(int)]); \
    })

int main() {
    int i = 12;
    std::cout << MAX_OF_N(1, 3, i, 6);
}

哦，是的，并且由于初始值设定项列表中存在潜在的变量表达式，我认为与此等效的内容（使用其自己的函数来避免 std::max_element）不会在C89。但我也不确定 C89 中是否有可变参数宏。

我认为这是绕过“只有一种类型”限制的东西。不过，它变得有点毛茸茸的：

#include <iostream>
#include <algorithm>

#define MAX_OF_N(x, ...) ({\
        typeof(x) ra[] = { (x), __VA_ARGS__ }; \
        *std::max_element(&ra[0], &ra[sizeof(ra)/sizeof(ra[0])]); \
    })

int main() {
    int i = 12;
    std::cout << MAX_OF_N(i + 1, 1, 3, 6, i);
}

You might consider this cheating, since it is not recursive and it doesn't do the work in the preprocessor. And it uses a GCC extension. And it only works for one type. It is, however, a variadic MAX_OF_N macro:

#include <iostream>
#include <algorithm>

#define MAX_OF_N(...) ({\
        int ra[] = { __VA_ARGS__ }; \
        *std::max_element(&ra[0], &ra[sizeof(ra) / sizeof(int)]); \
    })

int main() {
    int i = 12;
    std::cout << MAX_OF_N(1, 3, i, 6);
}

Oh yes, and because of the potential variable expression in the initializer list, I don't think that an equivalent of this (using its own function to avoid std::max_element) would work in C89. But I'm not sure variadic macros are in C89 either.

Here's something that I think gets around the "only one type" restriction. It's getting a bit hairy, though:

#include <iostream>
#include <algorithm>

#define MAX_OF_N(x, ...) ({\
        typeof(x) ra[] = { (x), __VA_ARGS__ }; \
        *std::max_element(&ra[0], &ra[sizeof(ra)/sizeof(ra[0])]); \
    })

int main() {
    int i = 12;
    std::cout << MAX_OF_N(i + 1, 1, 3, 6, i);
}

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尤怨 2024-07-26 08:14:40

不，因为预处理器只对文件进行一次“滑动”。没有办法让它递归地定义宏。

我见过的唯一执行类似操作的代码不是可变参数，而是使用用户必须传递的默认值：

x = MAX_OF_8 (a, b, -1, -1, -1, -1, -1, -1)

假设所有值都是非负的。

内联函数至少应该为 C++ 提供相同的功能。正如您所说，最好留给具有类似于 printf() 的可变参数的函数。

No, because the preprocessor only takes one "swipe" at the file. There's no way to get it to recursively define macros.

The only code that I've seen do something like this was not variadic, but used default values the user had to pass:

x = MAX_OF_8 (a, b, -1, -1, -1, -1, -1, -1)

assuming all values were non-negative.

Inline functions should give you the same for C++ at least. As you state, it's probably better left to a function with variable arguments similar to printf().

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若水微香 2024-07-26 08:14:40

我认为，即使您可以递归地扩展宏，您的方法在效率方面也会存在一个小问题...当扩展宏时，如果 MAX_OF_[N-1] 为更大，那么你必须从头开始重新评估它。

这是一个愚蠢而愚蠢的答案，可能没有人会喜欢 xD

文件“source.c”

#include "my_macros.h"
...

文件“Makefile”

myprogram: source.c my_macros.h
 gcc source.c -o myprogram

my_macros.h: make_macros.py
 python make_macros.py > my_macros.h

文件“make_macros.py”

def split(l):
    n = len(l)
    return l[:n/2], l[n/2:]

def gen_param_seq(n):
    return [chr(i + ord("A")) for i in range(n)]

def make_max(a, b):
    if len(a) == 1:
        parta = "("+a[0]+")"
    else:
        parta = make_max(*split(a))

    if len(b) == 1:
        partb = "("+b[0]+")"
    else:
        partb = make_max(*split(b))

    return "("+parta +">"+partb+"?"+parta+":"+partb+")"

for i in range(2, 9):
    p = gen_param_seq(i)
    print "#define MAX_"+str(i)+"("+", ".join(p)+") "+make_max(*split(p))

然后你将定义那些漂亮的宏：

#define MAX_2(A, B) ((A)>(B)?(A):(B))
#define MAX_3(A, B, C) ((A)>((B)>(C)?(B):(C))?(A):((B)>(C)?(B):(C)))
#define MAX_4(A, B, C, D) (((A)>(B)?(A):(B))>((C)>(D)?(C):(D))?((A)>(B)?(A):(B)):((C)>(D)?(C):(D)))
#define MAX_5(A, B, C, D, E) (((A)>(B)?(A):(B))>((C)>((D)>(E)?(D):(E))?(C):((D)>(E)?(D):(E)))?((A)>(B)?(A):(B)):((C)>((D)>(E)?(D):(E))?(C):((D)>(E)?(D):(E))))
#define MAX_6(A, B, C, D, E, F) (((A)>((B)>(C)?(B):(C))?(A):((B)>(C)?(B):(C)))>((D)>((E)>(F)?(E):(F))?(D):((E)>(F)?(E):(F)))?((A)>((B)>(C)?(B):(C))?(A):((B)>(C)?(B):(C))):((D)>((E)>(F)?(E):(F))?(D):((E)>(F)?(E):(F))))
#define MAX_7(A, B, C, D, E, F, G) (((A)>((B)>(C)?(B):(C))?(A):((B)>(C)?(B):(C)))>(((D)>(E)?(D):(E))>((F)>(G)?(F):(G))?((D)>(E)?(D):(E)):((F)>(G)?(F):(G)))?((A)>((B)>(C)?(B):(C))?(A):((B)>(C)?(B):(C))):(((D)>(E)?(D):(E))>((F)>(G)?(F):(G))?((D)>(E)?(D):(E)):((F)>(G)?(F):(G))))
#define MAX_8(A, B, C, D, E, F, G, H) ((((A)>(B)?(A):(B))>((C)>(D)?(C):(D))?((A)>(B)?(A):(B)):((C)>(D)?(C):(D)))>(((E)>(F)?(E):(F))>((G)>(H)?(G):(H))?((E)>(F)?(E):(F)):((G)>(H)?(G):(H)))?(((A)>(B)?(A):(B))>((C)>(D)?(C):(D))?((A)>(B)?(A):(B)):((C)>(D)?(C):(D))):(((E)>(F)?(E):(F))>((G)>(H)?(G):(H))?((E)>(F)?(E):(F)):((G)>(H)?(G):(H))))

而最好的事情是......有效^_^

I think that, even if you could expand macros recursively, there would be one little problem with your approach in terms of efficiency... when the macros are expanded, if the MAX_OF_[N-1] is greater, then you have to evaluate it again from scratch.

Here is a silly and stupid answer that probably no one will like xD

file "source.c"

#include "my_macros.h"
...

file "Makefile"

myprogram: source.c my_macros.h
 gcc source.c -o myprogram

my_macros.h: make_macros.py
 python make_macros.py > my_macros.h

file "make_macros.py"

def split(l):
    n = len(l)
    return l[:n/2], l[n/2:]

def gen_param_seq(n):
    return [chr(i + ord("A")) for i in range(n)]

def make_max(a, b):
    if len(a) == 1:
        parta = "("+a[0]+")"
    else:
        parta = make_max(*split(a))

    if len(b) == 1:
        partb = "("+b[0]+")"
    else:
        partb = make_max(*split(b))

    return "("+parta +">"+partb+"?"+parta+":"+partb+")"

for i in range(2, 9):
    p = gen_param_seq(i)
    print "#define MAX_"+str(i)+"("+", ".join(p)+") "+make_max(*split(p))

then you'll have those pretty macros defined:

#define MAX_2(A, B) ((A)>(B)?(A):(B))
#define MAX_3(A, B, C) ((A)>((B)>(C)?(B):(C))?(A):((B)>(C)?(B):(C)))
#define MAX_4(A, B, C, D) (((A)>(B)?(A):(B))>((C)>(D)?(C):(D))?((A)>(B)?(A):(B)):((C)>(D)?(C):(D)))
#define MAX_5(A, B, C, D, E) (((A)>(B)?(A):(B))>((C)>((D)>(E)?(D):(E))?(C):((D)>(E)?(D):(E)))?((A)>(B)?(A):(B)):((C)>((D)>(E)?(D):(E))?(C):((D)>(E)?(D):(E))))
#define MAX_6(A, B, C, D, E, F) (((A)>((B)>(C)?(B):(C))?(A):((B)>(C)?(B):(C)))>((D)>((E)>(F)?(E):(F))?(D):((E)>(F)?(E):(F)))?((A)>((B)>(C)?(B):(C))?(A):((B)>(C)?(B):(C))):((D)>((E)>(F)?(E):(F))?(D):((E)>(F)?(E):(F))))
#define MAX_7(A, B, C, D, E, F, G) (((A)>((B)>(C)?(B):(C))?(A):((B)>(C)?(B):(C)))>(((D)>(E)?(D):(E))>((F)>(G)?(F):(G))?((D)>(E)?(D):(E)):((F)>(G)?(F):(G)))?((A)>((B)>(C)?(B):(C))?(A):((B)>(C)?(B):(C))):(((D)>(E)?(D):(E))>((F)>(G)?(F):(G))?((D)>(E)?(D):(E)):((F)>(G)?(F):(G))))
#define MAX_8(A, B, C, D, E, F, G, H) ((((A)>(B)?(A):(B))>((C)>(D)?(C):(D))?((A)>(B)?(A):(B)):((C)>(D)?(C):(D)))>(((E)>(F)?(E):(F))>((G)>(H)?(G):(H))?((E)>(F)?(E):(F)):((G)>(H)?(G):(H)))?(((A)>(B)?(A):(B))>((C)>(D)?(C):(D))?((A)>(B)?(A):(B)):((C)>(D)?(C):(D))):(((E)>(F)?(E):(F))>((G)>(H)?(G):(H))?((E)>(F)?(E):(F)):((G)>(H)?(G):(H))))

and the best thing about it is that... it works ^_^

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顾冷 2024-07-26 08:14:40

如果您打算使用 C++ 走这条路，请查看模板元编程< /a>. 它并不漂亮，它可能无法解决您的确切问题，但它可以处理递归。

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囚我心虐我身 2024-07-26 08:14:40

首先，宏不会递归扩展。尽管如此，通过为每个递归级别创建一个宏，然后推导递归级别，宏可以具有可重入性。然而，所有这些重复和推论递归都是由 Boost.Preprocessor 库。因此，您可以使用高阶折叠宏来计算最大值：

#define MAX_EACH(s, x, y) BOOST_PP_IF(BOOST_PP_GREATER_EQUAL(x, y), x, y)
#define MAX(...) BOOST_PP_SEQ_FOLD_LEFT(MAX_EACH, 0, BOOST_PP_VARIADIC_TO_SEQ(__VA_ARGS__)) 

MAX(3, 6, 8) //Outputs 8
MAX(4, 5, 9, 2) //Outputs 9

现在，这将理解 0-256 之间的字面数字。它不适用于 C++ 变量或表达式，因为 C 预处理器不理解 C++。它只是纯文本替换。但是 C++ 提供了一种称为“函数”的功能，该功能适用于 C++ 表达式，您可以使用它来计算最大值。

template<class T>
T max(T x, T y)
{
    return x > y ? x : y;
}

template<class X, class... T>
auto max(X x, T ... args) -> decltype(max(x, max(args...)))
{
    return max(x, max(args...));
}

现在，上面的代码确实需要 C++11 编译器。如果您使用 C++03，则可以创建函数的多个重载以模拟可变参数。此外，我们可以使用预处理器为我们生成这些重复的代码（这就是它的用途）。所以在C++03中，你可以这样写：

template<class T>
T max(T x, T y)
{
    return x > y ? x : y;
}

#define MAX_FUNCTION(z, n, data) \
template<class T> \
T max(T x, BOOST_PP_ENUM_PARAMS(n, T x)) \
{ \
    return max(x, max(BOOST_PP_ENUM_PARAMS(n, x)));\
}

BOOST_PP_REPEAT_FROM_TO(2, 64, MAX_FUNCTION, ~)

First, macros don't expand recusrsively. Although, macros can have reentrance by creating a macro for each recursion level and then deducing the recursion level. However, all this repetition and deducing recursion, is taken care of by the Boost.Preprocessor library. You can therefore use the higher order fold macro to calculate the max:

#define MAX_EACH(s, x, y) BOOST_PP_IF(BOOST_PP_GREATER_EQUAL(x, y), x, y)
#define MAX(...) BOOST_PP_SEQ_FOLD_LEFT(MAX_EACH, 0, BOOST_PP_VARIADIC_TO_SEQ(__VA_ARGS__)) 

MAX(3, 6, 8) //Outputs 8
MAX(4, 5, 9, 2) //Outputs 9

Now, this will understand literal numbers between 0-256. It wont work on C++ variables or expression, because the C preprocessor doesn't understand C++. Its just pure text replacement. But C++ provides a feature called a "function" that will work on C++ expressions, and you can use it to calculate the max value.

template<class T>
T max(T x, T y)
{
    return x > y ? x : y;
}

template<class X, class... T>
auto max(X x, T ... args) -> decltype(max(x, max(args...)))
{
    return max(x, max(args...));
}

Now, the code above does require a C++11 compiler. If you are using C++03, you can create multiple overloads of the function in order to simulate variadic parameters. Furthermore, we can use the preprocessor to generate this repetitive code for us(thats what it is there for). So in C++03, you can write this:

template<class T>
T max(T x, T y)
{
    return x > y ? x : y;
}

#define MAX_FUNCTION(z, n, data) \
template<class T> \
T max(T x, BOOST_PP_ENUM_PARAMS(n, T x)) \
{ \
    return max(x, max(BOOST_PP_ENUM_PARAMS(n, x)));\
}

BOOST_PP_REPEAT_FROM_TO(2, 64, MAX_FUNCTION, ~)

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