通过循环（字符* C，...） [英] loop through (char *c, ...)

问题描述

我是新的C，和我通过常规的参数需要循环：

I'm new to C, and I need to loop through arguments of a routine:

void doSmth(char *c, ...) { //how to print all the elements here? }

由于我来自爪哇，这是一个相当新的给我，我不知道该怎么做这在C？

Since I come from Java, this is quite new to me, and I have no idea how to do this in C?

在此先感谢

推荐答案

由于您的函数的声明是这样的：

Because your function declaration is like:

无效doSmth（字符* C，...）;

您需要什么叫可变数量的参数功能，可以读取： 9.9。可变的参数个数一个很好的和作文教程

What you needs is called variable number of argument functions, you can read from : 9.9. Variable numbers of arguments a good and essay tutorial

例子code。与功能doSmth（）的第4步，阅读评论的：

Example code with function doSmth() its 4 steps, read comments:

//Step1: Need necessary header file
#include <stdarg.h>     
void doSmth( char* c, ...){   
    va_list ap;  // vlist variable
    int n;       // number 
    int i;
    float f;     

   //print fix numbers of arguments
   printf(" C: %s", c);

   //Step2: To initialize `ap` using right-most argument that is `c` 
    va_start(ap, c); 

   //Step3: Now access vlist `ap`  elements using va_arg()
     n = va_arg(ap, int); //first value in my list gives number of ele in list 
     while(n--){
       i = va_arg(ap, int);
       f = (float)va_arg(ap, double); //notice type, and typecast  
       printf("\n %d %f \n", i, f);
    }

    //Step4: Now work done, we should reset pointer to NULL
    va_end(ap); 
}
int main(){
    printf("call for 2");
    doSmth("C-string",  2,    3,   6.7f,  5,   5.5f);
        //              ^ this is `n` like count in variable list 
    printf("\ncall for 3");
    doSmth("CC-string", 3,  -12, -12.7f,-14, -14.4f, -67, -0.67f);
         //             ^ this is `n` like count in variable list
    return 1;
}

运行它，如：

:~$ ./a.out 
call for 2 C: C-string
 3 6.700000 
 5 5.500000 

call for 3 C: CC-string
 -12 -12.700000 
 -14 -14.400000 
 -67 -0.670000 

<子>在C语言中的东西其实是参数后跟可变的参数个数固定数量

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