将字符添加到字符串中 [英] adding chars to a string

问题描述

这可能很简单，当你从未做过指针并习惯于使用Delphi或Visual Basic中的
luxery字符串时，C可以尽管如此。我想要做的就是将字符串添加到字符串中。我看了一下string.h文件，但是

我没有找到那种功能（那个字符串库超过7年b
岁）所以我决定把这个功能写在自己身上。它似乎工作

虽然当我试图在这些动作之间做printf时我很奇怪

消息。无论如何，现在我正在尝试将字符添加到2个字符串。第一个字符串

是可以的，但第二个字符串似乎已经被淹没了。这是我的代码：


/ *将1个字符添加到字符串中的函数* /


char * strAddChar（char * s1，char c）{

char * s;

s = s1;

s = s + strlen（s1）; //写的位置

* s = c; //这应该在null char

的地方首先

//是。

s ++;

* s = 0; //添加一个新的空字符串

return（s1）;

} // strAddChar


/ *这就是我使用函数* /

main（）{

char * res1 ="" ;; //空字符串1

char * res2 ="" ;; //空字符串2

strAddChar（res1，'a''）;

strAddChar（res1，''b''）;

strAddChar（res1，''c''）;

strAddChar（res2，''1''）;

strAddChar（res2，'''''）;

strAddChar（res2，''3''）;

printf（"％s \ n"，res1）; // test

printf（"％s \ n"，res2）;


好​​的，输出应为res1 =" abc"和res2 =123但这是我的结果：

res1 =" abc" res2 =" bc123"

" bc"进入res2 ??如何修复这个功能？


问候，

瑞克

Hi,

This is probably simple byt when you never did pointers and being used to
luxery strings like in Delphi or Visual Basic, C can get though. What I''m
trying to do is to add chars to a string. I looked in the string.h file but
I didn''t find that kind of function (that string library is more than 7
years old) so I decided to write that function on myself. It seems to work
although when I''m trying to do printf''s between those actions I get strange
messages. Anyway, now I''m trying to add chars to 2 strings. The first string
is ok but the second one seems to get overwrited. Here''s my code :

/* the function to add 1 char to a string */

char *strAddChar(char *s1, char c){
char *s;
s = s1;
s = s + strlen(s1); // the position to write
*s = c; // this should be on the place where the null char
first
// was.
s++;
*s = 0; // add a new null string
return(s1);
} // strAddChar

/* this is how I use the function */
main(){
char *res1 = ""; // empty string 1
char *res2 = ""; // empty string 2
strAddChar( res1, ''a'' );
strAddChar( res1, ''b'' );
strAddChar( res1, ''c'' );
strAddChar( res2, ''1'' );
strAddChar( res2, ''2'' );
strAddChar( res2, ''3'' );
printf( "%s\n",res1 ); // test
printf( "%s\n",res2 );

Ok, the output should be res1="abc" and res2="123" but this is my result :
res1="abc" res2="bc123"
How does "bc" come in res2?? How to fix the function?

Greetings,
Rick

推荐答案

在文章< 3f *********************** @ news.xs4all.nl>中，Rick写道：

In article <3f***********************@news.xs4all.nl>, Rick wrote:

<当你从未做过指针并习惯于像Delphi或Visual Basic这样的luxery字符串时，这可能很简单，C可以得到。我试图做的是将字符串添加到字符串中。

Hi,

This is probably simple byt when you never did pointers and being used to
luxery strings like in Delphi or Visual Basic, C can get though. What I''m
trying to do is to add chars to a string.

您是否调查了strcat（）和strncat（）？

-

Andreas K？h？ri

Did you investigate strcat() and strncat()?
--
Andreas K?h?ri

要添加，我认为result2会被result1以某种方式覆盖。当那些2

被初始化时，它们会自动获得一个内存地址，而且当我将结果添加到result1时，似乎

，结果2不会进一步移动。结果1

可能只获得1个字节的内存，因此在添加第二个字符时，它的数量为b
。结果2上的和contineus。


如果这是真的，那将是非常危险的，因为我可以覆盖程序中的大部分内容。通常你可以使用数组和malloc它们，但在我的情况下，我真的不知道结果有多大。在构建

结果字符串时，我需要至少1个子字符串用于其他东西，该字符串有

同样的问题，我无法计算它有多大，它必须动态增长

。怎么处理这个？


问候，

Rick

To add, I think result2 get''s overwritten by result1 somehow. When those 2
got initialized they get a memory adress automatically and it seems that
when I''m adding chars to result1, result2 won''t move further. Result1
probably only get 1 byte of memory so when adding the second char it''s "out
of bounds" and contineus on result2.

If this is true, that would be quiete dangerous because I could overwrite a
big part of the program. Normally you can use arrays and malloc them but in
my case, I really don''t know how big the result becomes. While building the
result string, I need at least 1 sub string for other stuff, that string has
the same problem, I can''t calculate how big it becomes, it has to grow
dynamically. How to handle this?

Greetings,
Rick

你的答案比光快！我只是试了但是同样的问题

（看到其他帖子，我解释了问题一点点）。或者

也许我使用的代码错了，这就是我用strncat做的：


char * res1 ="" ;;

char * res2 ="" ;;

strncat（res1，" a"，1）;

strncat（res1，" b"， 1）;

strncat（res1，" c"，1）;

strncat（res2，" 1"，1）;

strncat（res2，" 2"，1）;

strncat（res2，" 3"，1）;

printf（" result1％s \ n"，res1）;

printf（" result2％s \ n"，res2）;


结果：res1 =" abc" RES2 = QUOT; bc123" ;.当使用strcat时一样。


问候，

Rick

You answered faster than the light! I just tried it but the same problem
(see that other post where I explained the problem a little bit more). Or
maybe I am using the code wrong, this is what I did with strncat:

char *res1 = "";
char *res2 = "";
strncat( res1, "a",1 );
strncat( res1, "b",1 );
strncat( res1, "c",1 );
strncat( res2, "1",1 );
strncat( res2, "2",1 );
strncat( res2, "3",1 );
printf( "result1 %s\n",res1 );
printf( "result2 %s\n",res2 );

result : res1 = "abc" res2="bc123". When using strcat the same.

Greetings,
Rick

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