在C中部分应用函数 [英] Partially applying a function in C

问题描述

我有以下代码:

/* Function that takes a list and applies "function" to every element */
struct list *iter(struct list *l, void (*function)(struct list *a));

我有一个函数 foo(arg1，arg2，struct list * a ，我想将其应用于列表中的每个元素.

And I have a function foo(arg1, arg2, struct list *a that I want to apply to every element of my list.

如何实现这种行为?在支持闭包或循环的编程语言中，我将对 arg1 和 arg2 部分地应用 foo ，但是在C语言中似乎是不可能的.

How can I achieve such a behaviour ? In a programming language that supports closures or currying I would have partially applied foo to arg1 and arg2, but it seems to be impossible in C.

是否可以让函数返回指向函数的指针 bar(struct list * a)= foo(arg1，arg2，a)吗?

Is it possible to have a function return a pointer to a function bar(struct list *a) = foo(arg1, arg2, a) ?

推荐答案

基于您的函数定义:

struct list *iter(struct list *l, void (*function)(struct list *a));

我假设您的列表由一系列 struct list 对象组成(与具有单独的Head和Node类型的列表相对).

I'm assuming that your list consists of a series of struct list objects (as opposed to lists which have a separate Head and Node type).

要使回调更通用，您需要提供一个额外的参数，调用方可以根据自己的需要定制该参数:

To make the callback more generally usable, you need to supply an extra argument which the caller can tailor to their own needs:

struct list *iter( struct list *l, void *arg, void (*function)(void *, struct list *));

在 iter 中，使用 function(arg，l);

然后您以这种方式调用 foo :

Then you invoke foo this way:

struct foo_args { arg1_t arg1; arg2_t arg2; };   // at file scope

void handler(void *args, struct list *node)    // at file scope
{
    struct foo_args *p_args = args;
    foo(p_args->arg1, p_args->arg2, node);
}

// in a function
struct foo_args args = { arg1, arg2 };
iter(l, &args, handler);

当前您不使用 function 的返回值，通常可以将其用作标志以指示迭代是否应中断(例如，由于处理此节点失败或发现了它的结果)正在寻找).

Currently you're not using function's return value, typically this can be used for a flag to indicate whether the iteration should break (e.g. because processing this node failed, or it found what it was looking for).

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