提取带有可变参数的函数 [英] extract function with variable arguments

问题描述

鉴于我有这个功能:

void A::IAmCool(int x, ...)
{
        va_list args; 
        va_start (args, x);   
        va_end (args);
}

如何将 var-args 从一个函数传递到另一个函数?我正在寻找这样的东西:

How do I pass var-args from one function to another? I'm looking for something like this:

void A::extractedFunction() /* this doesn't work */
{
        va_list args; 
        va_start (args, ?????);   
        va_end (args);
}

void A::IAmCool(int x, ...)
{
        extractedFunction();
}

这甚至可能吗?我试过使函数内联，但这不起作用.

Is this even possible? I have tried making the function inline but that doesn't work.

推荐答案

通常的模式是使用 valist 来实现您的主要主力函数，而可变参数函数仅作为装饰".然后就可以直接从第三方调用站点使用 main 函数了.示例:

The usual pattern is to implement your main workhorse function taking a valist, and the variadic function only as "decoration". Then you can use the main function directly from third-party call sites. Example:

#include <cstdarg>

int vgizmo(int a, std::va_list ap)
{
    // main implementation here!
}

int gizmo(int a, ...)  // interface function
{
    std::va_list ap;
    va_start(ap, a);
    int r = vgizmo(a, ap);
    va_end(ap);
    return r;
}

void some_other_stuff(bool q, char const * fmt, ...)
{
    std::va_list ap;
    va_start(ap, fmt);

    // ...

    int b = vgizmo(x, ap);   // third parties use vgizmo directly

    // ...

    va_end(ap);
}

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