C语言 C 字符到字符串（将字符传递给 strcat()）

C char to string (passing char to strcat())

提问by frx08

my problem is in convert a char to string i have to pass to strcat() a char to append to a string, how can i do? thanks!

我的问题是将字符转换为字符串我必须传递给 strcat() 一个字符以附加到字符串，我该怎么办？谢谢！

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

char *asd(char* in, char *out){
    while(*in){
        strcat(out, *in); // <-- err arg 2 makes pointer from integer without a cast
        *in++;
    }
    return out;
}

int main(){
    char st[] = "text";
    char ok[200];
    asd(st, ok);
    printf("%s", ok);
    return 0;
}

采纳答案by legends2k

Since okis pointing to an uninitialized array of characters, it'll all be garbage values, so where the concatenation (by strcat) will start is unknown. Also strcattakes a C-string (i.e. an array of characters which is terminated by a '\0' character). Giving char a[200] = ""will give you a[0] = '\0', then a[1] to a[199] set to 0.

由于ok指向一个未初始化的字符数组，所以它都是垃圾值，所以连接 (by strcat) 将从哪里开始是未知的。还strcat需要一个 C 字符串（即以 '\0' 字符结尾的字符数组）。给予char a[200] = ""会给你 a[0] = '\0'，然后 a[1] 到 a[199] 设置为 0。

Edit:(added the corrected version of the code)

编辑：（添加了代码的更正版本）

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

char *asd(char* in, char *out)
{

/*
    It is incorrect to pass `*in` since it'll give only the character pointed to 
    by `in`; passing `in` will give the starting address of the array to strcat
 */

    strcat(out, in);
    return out;
}

int main(){
    char st[] = "text";
    char ok[200] = "somevalue"; /* 's', 'o', 'm', 'e', 'v', 'a', 'l', 'u', 'e', '
char *asd(char* in, char *out)
{
    char *end = out + strlen(out);

    do
    {
        *end++ = *in;

    } while(*in++);

    return out;
}
' */
    asd(st, ok);
    printf("%s", ok);
    return 0;
}

回答by AndiDog

strcatwill not append single characters. Instead it takes a const char*(a full C-style string) which is appended to the string in the first parameter. So your function should read something like:

strcat不会附加单个字符。相反，它需要一个const char*（完整的 C 样式字符串）附加到第一个参数中的字符串。所以你的函数应该是这样的：

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

int main(){
    char st[] = "text";
    char ok[200];
    ok[0] = '
char* ctos(char c)
{
    char s[2];
    sprintf(s, "%c##代码##", c);
    return s;
}
'; /* OR
    memset(ok, 0, sizeof(ok));
    */
    strcat(ok, st);
    printf("%s", ok);
    return 0;
}

The do-while loop will include the zero-terminator which is necessary at the end of C-style strings. Make sure that your out string is initialized with a zero-terminator at the end or this example will fail.

do-while 循环将包括零终止符，这是 C 样式字符串末尾所必需的。确保您的输出字符串在末尾使用零终止符进行初始化，否则此示例将失败。

And as an aside: Think about what *in++;does. It will increment inand dereference it, which is the very same as in++, so the *is useless.

顺便说一句：想想是什么*in++;。它将增加in和取消引用它，这与 非常相同in++，因此*是无用的。

回答by t0mm13b

To look at your code, I can make a couple of pointers in relation to it, this is not a criticism, take this with a pinch of salt that will enable you to be a better C programmer:

为了查看您的代码，我可以提出一些与它相关的指针，这不是批评，请稍加保留，这将使您成为更好的 C 程序员：

• No function prototype.
• Incorrect usage of pointers
• Dealing with the strcatfunction is used incorrectly.
• Overdoing it - no need for the asdfunction itself!
• Usage of dealing with variables notably chararray that is not properly initialized.

• 没有函数原型。
• 指针使用不当
• 处理strcat函数使用不当。
• 过度使用 - 不需要asd函数本身！
• 处理变量的用法，特别char是未正确初始化的数组。

##代码##

Hope this helps, Best regards, Tom.

希望这会有所帮助，最好的问候，汤姆。

回答by Joseph

To convert a character to a (null terminated) string you could simply do:

要将字符转换为（空终止）字符串，您可以简单地执行以下操作：

##代码##

Working example: http://ideone.com/Cfav3e

工作示例：http: //ideone.com/Cfav3e