c - 获取 argv[0] 中 n 和 argv 长度之间的子字符串

Question

我正在尝试获取 argv[0] 的子字符串并将其输出到另一个字符串。我已经得到了我想要的职位，我只是不知道如何从其他答案中做到这一点。这是我到目前为止所得到的:

#include "echo.h"
#include "whoami.h"
#include <string.h>

int main(int argc, char *argv[])
{
    // applet_length: Length of first arg filename excluding things like "./"
    int applet_length = sizeof(argv[0]);
    // real_length: Full length of first argument
    int real_length = 0;
    while ((argv[0])[real_length])
        real_length++;
    int start = real_length - applet_length;
}

start 是我想要开始子字符串的点，real_length 是 argv[0] 的总长度。我该如何去做呢？

我这样做是为了制作一个像 BusyBox 这样的 coreutils 可执行文件，您可以通过调用带有其名称的符号链接(symbolic link)来运行小程序。

例如argv[0] 为 "test1234"，起始位置为 4，real_length 为 8，输出为 “1234”

<小时/>

我最终得到:

#include "echo.h"
#include "whoami.h"
// etc.
#include <string.h>
#include <libgen.h>

int main(int argc, char *argv[])
{
    char *applet = basename(argv[0]);
    if (!strcmp(applet, "echo"))
        echo(argc, argv);
    else if (!strcmp(applet, "whoami"))
        whoami(argc, argv);
    // etc.
    return 0;
}

Answer 1

这是一个带有注释的程序，解释了如何做你想做的事情。希望它能帮助您学习如何解决此类问题。

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

/* Use if strdup() is unavailable.  Caller must free the memory returned. */
char *dup_str(const char *s)
{
    size_t n = strlen(s) + 1;
    char *r;
    if ((r = malloc(n)) == NULL) {
        return NULL;
    }
    strcpy(r, s);
    return r;
}

/* Use if POSIX basename() is unavailable */
char *base_name(char *s)
{
    char *start;

    /* Find the last '/', and move past it if there is one.  Otherwise return
       a copy of the whole string. */
    /* strrchr() finds the last place where the given character is in a given
       string.  Returns NULL if not found. */
    if ((start = strrchr(s, '/')) == NULL) {
        start = s;
    } else {
        ++start;
    }
    /* If you don't want to do anything interesting with the returned value,
       i.e., if you just want to print it for example, you can just return
       'start' here (and then you don't need dup_str(), or to free
       the result). */
    return dup_str(start);
}

/* test */
int main(int argc, char *argv[])
{
    char *b = base_name(argv[0]);

    if (b) {
        printf("%s\n", b);
    }
    /* Don't free if you removed dup_str() call above */
    free(b);

    return EXIT_SUCCESS;
}