arrays character c command-line-arguments

C 参数字符数组

发布于 2025-01-03 18:56:36 字数 454 浏览 0 评论 0原文

我想传递参数并将它们输出为字符或字符串数组。做到这一点最简单的方法是什么？一旦我将它们作为字符数组，我希望能够操作该数组。我能够传递它们并使用字符指针遍历字符串，但我想使用字符数组。有什么简单的方法可以做到这一点吗？我对 C 还很陌生，这是我到目前为止所掌握的。

    if(argc != 3){
            printf("incorrect number of arguments, program needs 3\n");
            exit(1);
    }

    char s1[20];
    s1 = (char*)argv[1];
    char s2[20];
    s2 = (char*)argv[2];

    printf("String1 is %s\n", s1);
    printf("String2 is %s\n", s2);

    exit(0);

原文

I would like to pass in arguments and output them as an array of characters or string. What is the easiest way to do this? Once I have them as an array of characters I want to be able to manipulate the array. I was able to pass them in and use a character pointer to traverse through the string but I'd like to use a char array. Is there any easy way to do this? Im pretty new to C, here is what I have so far.

    if(argc != 3){
            printf("incorrect number of arguments, program needs 3\n");
            exit(1);
    }

    char s1[20];
    s1 = (char*)argv[1];
    char s2[20];
    s2 = (char*)argv[2];

    printf("String1 is %s\n", s1);
    printf("String2 is %s\n", s2);

    exit(0);

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没有心的人 2025-01-10 18:56:37

argv 数组指向的字符串是可以修改的，你可以只修改argv 数组指向的字符串，不需要复制它们。

从C标准来看：

(C99, 5.1.2.2.1p2) “参数 argc 和 argv 以及 argv 数组指向的字符串应可由程序修改，并在程序启动和程序终止之间保留其最后存储的值。”< /p>

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黯然 2025-01-10 18:56:37

您可以直接直接打印 argv 的值：

printf("String1 is %s\n", argv[1]);
printf("String2 is %s\n", argv[2]);

如果您需要修改字符串，请按以下方法操作

char *newString1 = malloc(strlen(argv[1]) + 1 +  charactersToAdd); // add one for null termination.
strcpy(newString1, argv[1]);

// manipulate newString1 however you choose
printf("String1 is %s\n", newString1);
free(newString1);

You can simply print the values of argv directly:

printf("String1 is %s\n", argv[1]);
printf("String2 is %s\n", argv[2]);

If you need to modify the strings, here is how you would do that

char *newString1 = malloc(strlen(argv[1]) + 1 +  charactersToAdd); // add one for null termination.
strcpy(newString1, argv[1]);

// manipulate newString1 however you choose
printf("String1 is %s\n", newString1);
free(newString1);

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潇烟暮雨 2025-01-10 18:56:37

argv[1] 和 argv[2] 已经是指向字符数组的指针，因此您可以直接使用它们。此外（请参阅@ouah 的帖子），字符串已经是可变的，因此您可能根本不需要除此之外的任何内容。

如果您确实想要一个作用域本地数组，则需要使用称为“可变长度数组”的 C99 功能，因为字符串长度仅在运行时已知：

#include <string.h>

int main(int argc, char * argv[])
{
    char s1[strlen(argv[1]) + 1];
    strncpy(s1, argv[1], sizeof s1);

    // ...
}

或者您可以malloc 足够的空间用于字符串的副本（例如 char * s1 = malloc(strlen(argv[1]) + 1);），但是这样你就没有本地毕竟数组（你可能会这样我们将使用原始字符串，除非您真的想复制）。

argv[1] and argv[2] are already pointers to character arrays, so you can use those directly. Moreover (see @ouah's post), the strings are already mutable, so you might not need anything beyond this at all.

If you really want to have a scope-local array, you need to use the C99 feature called "variable-length arrays", since the string lengths are only known at runtime:

#include <string.h>

int main(int argc, char * argv[])
{
    char s1[strlen(argv[1]) + 1];
    strncpy(s1, argv[1], sizeof s1);

    // ...
}

Alternatively you could malloc enough space for a copy of the string (say char * s1 = malloc(strlen(argv[1]) + 1);), but then you wouldn't have a local array after all (and you might as well use the original strings, unless you really wanted to make a copy).

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旧时模样 2025-01-10 18:56:37

argv 已经是 char* （实际上是数组的数组），但是，它们被标记为 const，如“不要修改”；）。你想要的实际上是复制它们，这样你就可以自由地操纵它们：

char str[40];
strcpy (str,argv[1]);

现在你可以按照你的意愿操纵 str 。

argv is already char* (actually an array of an array), however, they are noted const as in "do not modify" ;). What you want is actually making a copy of them so you can manipulate them freely: