[英]C char to string (passing char to strcat())
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? 我的问题是将char转换为字符串我必须传递给strcat()一个字符串附加到字符串,我该怎么办? thanks! 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;
}
Since ok
is pointing to an uninitialized array of characters, it'll all be garbage values, so where the concatenation (by strcat
) will start is unknown. 由于ok
指向未初始化的字符数组,因此它们都是垃圾值,因此串联(通过strcat
)将在何处开始是未知的。 Also strcat
takes a C-string (ie an array of characters which is terminated by a '\\0' character). strcat
采用C字符串(即由'\\ 0'字符终止的字符数组)。 Giving char a[200] = ""
will give you a[0] = '\\0', then a[1] to a[199] set to 0. 给char a[200] = ""
会给你一个[0] ='\\ 0',然后[1]到[199]设置为0。
Edit: (added the corrected version of the code) 编辑:(添加了更正的代码版本)
#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', '\0' */
asd(st, ok);
printf("%s", ok);
return 0;
}
strcat
will not append single characters. strcat
不会附加单个字符。 Instead it takes a const char*
(a full C-style string) which is appended to the string in the first parameter. 相反,它需要一个const char*
(一个完整的C风格的字符串),它附加在第一个参数的字符串中。 So your function should read something like: 所以你的函数应该是这样的:
char *asd(char* in, char *out)
{
char *end = out + strlen(out);
do
{
*end++ = *in;
} while(*in++);
return out;
}
The do-while loop will include the zero-terminator which is necessary at the end of C-style strings. do-while循环将包括在C样式字符串结尾处必需的零终止符。 Make sure that your out string is initialized with a zero-terminator at the end or this example will fail. 确保您的out字符串在结尾处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++;
除此之外:想想*in++;
does. 确实。 It will increment in
and dereference it, which is the very same as in++
, so the *
is useless. 这将增加in
和取消对它的引用,这是非常相同in++
,所以*
是没用的。
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: 为了查看你的代码,我可以提出一些与之相关的指示,这不是一个批评,用一点盐来实现,这将使你成为一个更好的C程序员:
strcat
function is used incorrectly. 处理strcat
函数使用不正确。 asd
function itself! 过度使用 - 不需要asd
功能本身! char
array that is not properly initialized. 处理变量的用法,特别是未正确初始化的char
数组。 #include <stdio.h> #include <string.h> int main(){ char st[] = "text"; char ok[200]; ok[0] = '\0'; /* OR memset(ok, 0, sizeof(ok)); */ strcat(ok, st); printf("%s", ok); return 0; }
Hope this helps, Best regards, Tom. 希望这会有所帮助,最好的问候,汤姆。
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:
char* ctos(char c)
{
char s[2];
sprintf(s, "%c\0", c);
return s;
}
Working example: http://ideone.com/Cfav3e 工作示例: http : //ideone.com/Cfav3e
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