以指针为参数的可变参数函数

逐案命名

程序应该打印给定数组(大小为 8)的前三个、五个和八个元素的总和。我已经能够main像这样函数中传递指针

代码:

#include <stdlib.h>
#include <stdio.h>
#include <stdarg.h>
int  sum (int count,int data[],...) {
    int res=0;
    for(int i=0;i<count;i++)
    {
        res=res+data[i];   
    }
    return res;
}
int main(){
    int elements[8]={11,2,33,5,8,48,2,-8};
    int three_elements;
    int five_elements;
    int eight_elements;
    three_elements=sum(3, &elements[0], &elements[1], &elements[2]);
    five_elements=sum(5, &elements[0], &elements[1], &elements[2], &elements[3], &elements[4]);
    eight_elements=sum(8, &elements[0], &elements[1], &elements[2], &elements[3], &elements[4],&elements[5], &elements[6], &elements[7]);
    printf("First Three elements sum:%d\n",three_elements);
    printf("First five elements sum:%d\n",five_elements);
    printf("First eight elements sum:%d",eight_elements);
    return 0;
}

输出:

前三元素和:46
前五元素和:59
前八元素和:101

如何通过将参数直接传递给函数来做到这一点sum我试图这样做,但编译器不断给我错误:

代码:

#include <stdlib.h>
#include <stdio.h>
#include <stdarg.h>
int  sum (int count,int* data[],...) {
    int res=0;
    for(int i=0;i<count;i++)
    {
        res=res+data[i];   
    }
    return res;
}
int main(){
    int elements[8]={11,2,33,5,8,48,2,-8};
    int three_elements;
    int five_elements;
    int eight_elements;
    three_elements=sum(3, elements[0], elements[1], elements[2]);
    five_elements=sum(5, elements[0], elements[1], elements[2], elements[3], elements[4]);
    eight_elements=sum(8, elements[0], elements[1], elements[2], elements[3], elements[4],elements[5], elements[6], elements[7]);
    printf("First Three elements sum:%d\n",three_elements);
    printf("First five elements sum:%d\n",five_elements);
    printf("First eight elements sum:%d",eight_elements);
    return 0;
}
安东尼·加弗莱尔

可变函数之和

看看男人

va_start() 宏初始化 ap 以供 va_arg() 和 va_end() 后续使用,并且必须首先调用。

va_arg() 宏扩展为一个表达式,该表达式具有调用中下一个参数的类型和值。参数 ap 是由 va_start() 初始化的 va_list ap。每次调用 va_arg() 都会修改 ap 以便下一次调用返回下一个参数。参数类型是指定的类型名称,以便可以通过在类型上添加 * 来简单地获得指向具有指定类型的对象的指针的类型。

没有指针,您可以通过以下方式进行操作:

#include <stdlib.h>
#include <stdio.h>
#include <stdarg.h>

int  sum (int len, int value, ...) {
    va_list args; // initialize va_list

    int res = value;
    va_start(args, value); // start va
    for (int i = 1; i < len; i++) // iterate over other arguments
        res += va_arg(args, int); // add the next argument to res
    va_end(args); // don't forget to end va
    return res;
}
int main(){
    int elements[8]={11,2,33,5,8,48,2,-8};
    int three_elements;
    int five_elements;
    int eight_elements;
    three_elements=sum(3, elements[0], elements[1], elements[2]);
    five_elements=sum(5, elements[0], elements[1], elements[2], elements[3], elements[4]);
    eight_elements=sum(8, elements[0], elements[1], elements[2], elements[3], elements[4],elements[5], elements[6], elements[7]);
    printf("First Three elements sum:%d\n",three_elements);
    printf("First five elements sum:%d\n",five_elements);
    printf("First eight elements sum:%d\n",eight_elements);
    return 0;
}

为了通过指针获得总和,您可以通过以下方式进行:

#include <stdlib.h>
#include <stdio.h>
#include <stdarg.h>

int  sum (int len, int *value, ...) {
    va_list args; // initialize va_list

    int res = *value;
    va_start(args, *value); // start va
    for (int i = 1; i < len; i++) // iterate over other arguments
        res += (int)*va_arg(args, int*); // add the next argument to res
    va_end(args); // don't forget to end va
    return res;
}
int main(){
    int elements[8]={11,2,33,5,8,48,2,-8};
    int three_elements;
    int five_elements;
    int eight_elements;
    three_elements=sum(3, &elements[0], &elements[1], &elements[2]);
    five_elements=sum(5, &elements[0], &elements[1], &elements[2], &elements[3], &elements[4]);
    eight_elements=sum(8, &elements[0], &elements[1], &elements[2], &elements[3], &elements[4], \
        &elements[5], &elements[6], &elements[7]);
    printf("First Three elements sum:%d\n",three_elements);
    printf("First five elements sum:%d\n",five_elements);
    printf("First eight elements sum:%d\n",eight_elements);
    return 0;
}

输出:

First Three elements sum:46
First five elements sum:59
First eight elements sum:101

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