C语言 将 argv[1] 存储到 char 变量

store argv[1] to an char variable

提问by yaylitzis

I pass a character to my program and I want to store this character to variable. For example I run my program like this ./a.out s file. Now I want to save the argv[1] (its the s) to a variable (lets say I define it like this char ch;. I saw this approach:

我将一个字符传递给我的程序，我想将此字符存储到变量中。例如，我像这样运行我的程序./a.out s file。现在我想将 argv[1] （它的 s）保存到一个变量中（假设我像这样定义它char ch;。我看到了这种方法：

ch = argv[1][0];

and it works. But I cant understand it. The array argv isnt a one dimensional array? if i remove the [0], I get a warning warning: assignment makes integer from pointer without a cast

它有效。但我无法理解。数组 argv 不是一维数组？如果我删除 [0]，我会收到警告warning: assignment makes integer from pointer without a cast

回答by trojanfoe

If you look at the declaration of main()you see that it's

如果你看一下你的声明，main()你会发现它是

int main(int argc, const char **argv);

or

或者

int main(int argc, const char *argv[]);

So argvis an array of const char *(i.e. character pointers or "C strings"). If you dereference argv[1]you'll get:

所以argv是阵列const char *（即，字符指针或“C串”）。如果您取消引用，argv[1]您将得到：

"s"

or:

或者：

{ 's' , '
's'
' }

and if you dereference argv[1][0], you'll get:

如果你取消引用argv[1][0]，你会得到：

const char *myarg = NULL;

int main(int argc, const char **argv) {
    if (argc != 2) {
        fprintf(stderr, "usage: myprog myarg\n");
        return 1;
    } else if (strlen(argv[1]) != 1) {
        fprintf(stderr, "Invalid argument '%s'\n", argv[1]);
        return 2;
    }

    myarg = argv[1];

    // Use argument as myarg[0] from now on

}

As a side note, there is no need to copy that character from argv[1], you could simply do:

作为旁注，无需从 复制该字符argv[1]，您可以简单地执行以下操作：

char* argv[]

回答by Yu Hao

argvis an array of strings, or say, an array of char *. So the type of argv[1]is char *, and the type of argv[1][0]is char.

argv是一个字符串数组，或者说，一个char *. 所以类型argv[1]是char *，类型argv[1][0]是char。

回答by David Heffernan

The typical declaration for argvis

的典型声明argv是

int main(int argc, char *argv[])

That is an array of char*. Now char*itself is, here, a pointer to a null-terminated array of char.

那是一个char*. 现在char*它本身就是一个指向空终止数组的指针char。

So, argv[1]is of type char*, which is an array. So you need another application of the []operator to get an element of that array, of type char.

所以，argv[1]是类型char*，它是一个数组。因此，您需要使用[]运算符的另一个应用程序来获取该数组的元素，类型为char。

回答by Juan Ramirez

Lets see mains' signature:

让我们看看电源的签名：

##代码##

No, argvis not a one-dimensional array. Its is a two-dimensional char array.

不，argv不是一维数组。它是一个二维字符数组。

1. argv[1]returns char*

2. argv[1][0]returns the first char in in argv[1]

1. argv[1]返回字符*

2. argv[1][0]返回 in 中的第一个字符 argv[1]