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没有 if 的语言？

发布于 2024-10-12 12:00:04 字数 48 浏览 5 评论 0原文

一位同事说他听说过一种没有“如果”概念的语言。这可能吗？如果是的话，它是什么语言？

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水晶透心 2024-10-19 12:00:04

除了 Prolog 之外，我不知道任何特定的语言，但我可以想到没有 if 语句 的语言可以工作的几种方式。事实上，您也不需要循环构造。显然，您需要某种条件分支和循环的方式。

例如，如果您具有以下功能：函数、函数参数上的 ML 样式模式匹配和尾部调用优化，您可以没有 if 或循环的程序。

foo () {
    for (i = 1 to 10) {
        if even(i) {
            print "even"
        }
    }
}

会变成类似

print_if_true (true) {
    print "even"
}
print_if_true (false) {}

foo_loop (11) {
}
foo_loop (n) {
    print_if_true(even(n))
    foo_loop(n+1)
}

foo () {
    foo_loop(1)
}

or 的类似 ML 语法：

foo => 
    let loop 11 => 0
              n => p_i_t(n), loop n + 1
    and p_i_t true => print "even"
                 _ => unit
    in
        loop 1
    end

当然，您仍然需要常用的比较运算符，然后您可以使用简单的 true/false 函数参数模式匹配而不是条件。或者您可以匹配任意值。或者该语言可以支持保护表达式，这些表达式基本上是确定函数重载是否有效的 if 语句。

上面的例子显然是人为的，没有 ifs/loops 的代码比原来的代码更难看、更难理解，但它演示了如何凑合。更多或不同的语言功能可能使编写没有 if/循环的干净程序成为可能。

另一种方法是这样的，如果 true == 1 且 false == 0。

[function(){else-clause}, function(){then-clause}][condition]()

也就是说，将 true 和 false 分支存储在列表或元组或任何可以通过 true 和 false 索引的内容中，然后使用结果以条件为索引，查找分支并调用函数。如果您的语言支持宏，则可以将传统条件翻译成这种格式。

Besides perhaps Prolog, I don't know of any specific languages, but I can think of a few ways a language without if statements may work. In fact, you don't need loop constructs either. You obviously needs some way of conditional branches and looping.

If, for example, you had the following features: functions, ML-style pattern matching on function arguments and tail-call optimization, you could program without if's or loops.

foo () {
    for (i = 1 to 10) {
        if even(i) {
            print "even"
        }
    }
}

would become soemthing like

print_if_true (true) {
    print "even"
}
print_if_true (false) {}

foo_loop (11) {
}
foo_loop (n) {
    print_if_true(even(n))
    foo_loop(n+1)
}

foo () {
    foo_loop(1)
}

or with ML-like syntax:

foo => 
    let loop 11 => 0
              n => p_i_t(n), loop n + 1
    and p_i_t true => print "even"
                 _ => unit
    in
        loop 1
    end

Of course, you still need the usual comparison operators and then you can use simple true/false function argument pattern matching instead of conditionals. Or you could match on arbitrary values. Or the language could support guard expressions, which are basically if statements which determine if a function overload is valid or not.

The example above is obviously contrived and the code without ifs/loops is a lot uglier and harder to understand than the original, but it demonstrates how you can make do. More or different language features may make it possible to write clean programs without if/loops.

Another way would be something like this, if true == 1 and false == 0.

[function(){else-clause}, function(){then-clause}][condition]()

That is, store the true and false branch in a list or tuple or whatever you have that can be indexed by true and false and then use the result of the condition as the index, look up the branch and call the function. If your language supports macros, it may be possibly to translate traditional conditionals into this format.

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何止钟意 2024-10-19 12:00:04

Smalltalk被认为是“真正的”面向对象语言，没有“if”语句，也没有“for”语句，没有“while”语句。还有其他例子（例如 Haskell），但这是一个很好的例子。
来源：没有 ifs

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一袭白衣梦中忆 2024-10-19 12:00:04

C++ 模板编程没有“if”构造，但通过模板专门化实现了图灵完备：

template <int N>
struct Factorial 
{
    enum { value = N * Factorial<N - 1>::value };
};

template <>
struct Factorial<0> 
{
    enum { value = 1 };
};

// Factorial<4>::value == 24
// Factorial<0>::value == 1

来自 Wikipedia 关于模板的文章元编程

C++ template programming does not have an 'if' construct but is Turing-complete via template specialization:

template <int N>
struct Factorial 
{
    enum { value = N * Factorial<N - 1>::value };
};

template <>
struct Factorial<0> 
{
    enum { value = 1 };
};

// Factorial<4>::value == 24
// Factorial<0>::value == 1

from Wikipedia's article on template metaprogramming

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