如何将 argv 分配给 C 中的指针变量？

Question

我正在尝试打印传递给 C 程序的参数，我想打印多次。下面是代码。你可以看到，我创建了一个 count 变量并在每个 while 循环之前设置为 argc 。是否有与 argv 类似的东西，以便它可以在 while 循环之前重置为 argv[0]？这是一个数组指针，我没有找到合适的方法。

#include <stdio.h>

int main(int argc, char const *argv[])
{
    int count = argc;
    while (count-- > 0)
    {
        printf("%s%s", *argv++, (count > 0) ? " " : "");
    }
    printf("\n");

    count = argc;
    while (count-- > 0)
    {
        // how to reset argv to point to argv[0] before calling *argv++
        printf((count > 0) ? "%s " : "%s", *argv++);
    }
    printf("\n");

    return 0;
}

Answer 1

你对argv的定义是错误的。来自 C11 标准 §5.1.2.2.1/2::

The parameters argc and argv and the strings pointed to by the argv array shall be modiﬁable by the program, and retain their last-stored values between program startup and program termination.

应该是char **argv或者char *argv[]

int main(int argc, char *argv[])
{
    int count = argc;

    char **savedArgv = argv;
    while (count--)
    {
        printf("%s%s", *argv++, (count > 0) ? " " : "");
    }
    printf("\n");

    count = argc;
    argv = savedArgv;
    while (count--)
    {
        // how to reset argv to point to argv[0] before calling *argv++
        printf((count > 0) ? "%s " : "%s", *argv++);
    }
    printf("\n");

    return 0;
}

https://godbolt.org/z/Me4Pq8

如何将 argv 分配给 C 中的指针变量？

How to assign argv to a pointer variable in C?

arrays

c

pointers

argv