传递函数的参数不确定数量和可变参数调用它 [英] Pass a function with undetermined number of arguments and call it with the variadic arguments

问题描述

我想创建一个接受与参数作为参数数目不详的函数的函数偷懒。我需要什么类型或强制转换做？

I'd like to create a function "lazy" which accepts a function with an undetermined number of arguments as parameter. What type do I need or which casts have to be done?

然后我想执行后面的功能评估说事儿。我怎么那么前通过我传递的参数给懒人功能来传递函数指针？

Then I want to execute that thing later in function "evaluate". How do I then pass the arguments I passed before to the "lazy" function to the passed function pointer?

有些code来说明我的问题：

Some code to illustrate my problem:

char *function_I_want_to_call(void *foo, type bar, ...);
// the arguments are unknown to lazy() and evaluate()

typedef struct {
    ??? func;
    va_list args;
} lazy_fcall;

void lazy(lazy_fcall *result, ??? func, ...) {
// which type do I need here?
    va_start(result->_args, fund);
    result->func = func;
}

void *evaluate(lazy_fcall *to_evaluate) {
    return to_evaluate->func(expand_args(to_evaluate->args));
    // what do I have to write there to expand the va_list? It's C, not C++11...
}

int main () {
    lazy_fcall lazy_store;
    lazy(&lazy_store, function_I_want_to_call, "argument_1", "argument_2");
    // ...
    printf("%s", (char *)evaluate(&lazy_store));
}

或者是这样的只是不可能的？其中其他的可能性存在吗？

Or is something like this just impossible? Which other possibilities exist?

推荐答案

您不能展开的va_list 成单独的参数。要调用函数必须能够采取的va_list 作为参数。例如参见的printf 与 vprintf 。

You can't expand a va_list into separate arguments. The function you want to call have to be able to take a va_list as argument. See e.g. printf versus vprintf.

另外，由CAF所指出的，你不能存储的va_list ，作为论据指出由它不会是有效的，一旦懒函数返回。尝试使用的va_list 将导致不确定的行为和各种古怪的。

Also, as noted by caf, you can't store a va_list, as the arguments "pointed" to by it will not be valid once the lazy function returns. Attempting to use the va_list will lead to undefined behavior and all kinds of weirdness.

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