C：传递可变数量的参数从一个函数到另一个 [英] C: Passing variable number of arguments from one function to another

问题描述

所以，这里的我面对现在的一个小问题 - >我试图写将接受一个char *的消息和参数变量数的函数。我的功能将修改的消息一点点，然后它会调用与消息和给定参数的printf。 Essentialy，我试着写类似的东西：

So, here's a small problem I'm facing right now -> I'm trying to write a function that will accept a char* message and a variable number of arguments. My function will modify the message a little, and then It'll call printf with the message and given parameters. Essentialy, I'm trying to write something like that:

void modifyAndPrintMessage(char* message,...){
    char* newMessage; //copy message.
    //Here I'm modifying the newMessage to be printed,and then I'd like to print it. 
    //passed args won't be changed in any way.

    printf(newMessage,...); //Of course, this won't work. Any ideas?
    fflush(stdout);

}

因此​​，任何人知道我应该怎么办做到这一点？我会非常感谢任何帮助：）

So, anybody knows what should I do to make it happen? I'd be most grateful for any help :)

推荐答案

您想使用可变参数...

You want to use varargs...

void modifyAndPrintMessage( char* message, ... )
{
    // do somehthing custom

    va_list args;
    va_start( args, message );

    vprintf( newMessage, args );

    va_end( args );
}

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