在 C 中使用 va_arg 搜索密钥

Question

The first number taken as argument is the key which is to be found in the numbers following it. My code ignores the first number and and stores the second number in key , and also fails to return a correct result when I add a zero at the beginning of the argument to fix choosing the second argument.

The code:

int search(int n, ...)
{
    va_list vl;
    va_start(vl,n);
    int key,flg=0;
    key=va_arg(vl,int);
    printf("key %d\n",key);
    for(int i=2;i<=n;i++)
    {
        int x=va_arg(vl,int);
        if(key==x)
        {
            printf("FOUND!\n");
            flg=1;
            break;
        }
    }    
    if(flg==0)  
        printf("NOT FOUND!\n");
    va_end(vl);
    return 0;
}

int main(int argc,char *argv[])
{
    if(argc<=1)
    {
        printf("No arguments passed!");
    }    
    else
    {
        int key=atoi(argv[1]);
        search(0,key,34,45,12,78,11,4,90,89,54,63,7,98,43);
    } 
    return 0;
}

Answer 1

This code is close to your code. I altered the error handling in main() — error messages should be reported on stderr , not stdout , and should end with a newline. The main() function loops over multiple arguments so a single run can test multiple values.

I also changed the call so that the number of arguments in the variable list is given as n , the first argument. There is no point in always returning 0 so I changed the function to return void (nothing). Alternatively, you should return flg so the calling code can tell whether the value was found or not. With the function returning no value, this yields the code:

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

static void search(int n, ...)
{
    va_list vl;
    va_start(vl, n);
    int key, flg = 0;
    key = va_arg(vl, int);
    printf("key %d\n", key);
    for (int i = 0; i < n; i++)
    {
        int x = va_arg(vl, int);
        if (key == x)
        {
            printf("FOUND!\n");
            flg = 1;
            break;
        }
    }
    if (flg == 0)
        printf("NOT FOUND!\n");
    va_end(vl);
}

int main(int argc, char *argv[])
{
    if (argc <= 1)
    {
        fprintf(stderr, "Usage: %s key\n", argv[0]);
        exit(EXIT_FAILURE);
    }
    for (int i = 1; i < argc; i++)
    {
        int key = atoi(argv[i]);
        search(13, key, 34, 45, 12, 78, 11, 4, 90, 89, 54, 63, 7, 98, 43);
    }
    return 0;
}

When run (source va71.c , program va71 ), I get:

$ va71 89 78 34 43 199 -1
key 89
FOUND!
key 78
FOUND!
key 34
FOUND!
key 43
FOUND!
key 199
NOT FOUND!
key -1
NOT FOUND!
$

which is what I'd expect.

Here's a reorganized version which reports the position where the key is found. There's no printing in the search function — keep I/O separate from the searching.

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

static int search(int n, int key, ...)
{
    va_list vl;
    va_start(vl, key);
    int flg = -1;
    for (int i = 0; i < n; i++)
    {
        int x = va_arg(vl, int);
        if (key == x)
        {
            flg = i;
            break;
        }
    }
    va_end(vl);
    return flg;
}

int main(int argc, char *argv[])
{
    if (argc <= 1)
    {
        fprintf(stderr, "Usage: %s key\n", argv[0]);
        exit(EXIT_FAILURE);
    }
    for (int i = 1; i < argc; i++)
    {
        int key = atoi(argv[i]);
        printf("key %d: ", key);
        int rv = search(13, key, 34, 45, 12, 78, 11, 4, 90, 89, 54, 63, 7, 98, 43);
        if (rv < 0)
            printf("not found\n");
        else
            printf("found at index %d\n", rv);
    }
    return 0;
}

Sample output:

$ va71 89 78 34 43 199 -1
key 89: found at index 7
key 78: found at index 3
key 34: found at index 0
key 43: found at index 12
key 199: not found
key -1: not found
$