填补空间 [英] fill space

问题描述

抱歉。我还有一个问题

我得到一个包含4个字符的字符串。我想把它放在一个26字符串的字符串中。怎么

我可以填充剩余字符的空间.. ??

是我需要做一个while循环来填充字符串的空间吗？


请帮忙！


谢谢

杰克

解决方案

" JC" < [e-mail]>在消息新闻中写道：bl ********* @ rain.i-cable.com ...

我得到一个包含4个字符的字符串。我想把它放在一个26字符串的字符串中。
如何填充剩余字符的空间.. ??
是我需要做一个while循环来填充字符串的空间吗？

不确定你想要什么，但是下面的程序可能很好

start。

#include< string.h>


int main（无效）

{

char small [] =" abc" ;; / *由4个字符组成的字符串* /

char large [26] ="" ;; / *字符串由26个字符组成* /


/ *将小字符串复制为大号，终止''\0''排除* /

memcpy（大，小，尺寸小 - 1）;

/ *将剩余的字符设置为'' - ''* /

memset（大+ sizeof小） - 1，'' - ''，sizeof large - sizeof small）;

/ *使用''\0''终止大号以使其成为字符串* /

large [sizeof large - 1] = 0;


/ * large现在包含字符串abc ---------------- ------" * /


返回0;

}

文章< bl *****[e-mail]> ;, JC写道：

抱歉。我还有一个问题
我有一个4字符串的字符串。我想把它放在一个26字符串的字符串中。怎么
我可以填充剩余字符的空间.. ??
是我需要做一个while循环填充字符串的空间？

#include< string.h>


/ * ... * /


char shorty [] =" FOO英寸; / *四个字符，包括终止* /

char冗长[26] = {0};


/ * ... * /


strcpy（冗长，短暂）;


shorty [3]的字符串终止符将被复制到

lenthy字符串，终止它[3]。你不需要

填写带有任何东西的字符串。


如果你*想要填充字符串的其余部分（元素

，索引4到25），那么memset （）会为你做这个：


memset（&（冗长[4]），'''''，21）;

冗长[ 25] =''\''';


for-loop将做同样的事情：


int i;

for（i = 4; i< 25; ++ i）{

longy [i] =''？'';

}

冗长[25] =''\ 0'';


字符串仍将在索引3处终止，除非

你用其他东西覆盖那个位置的''\ 0''。


-

Andreas K？h？ri

谢谢。

这是非常有用的编码。

" dis" < di*@hotmail.com>在消息中写道

news：bl ********** @ news3.tilbu1.nb.home.nl ...

" JC" < [e-mail]>在消息新闻中写道：bl ********* @ rain.i-cable.com ...

我得到一个包含4个字符的字符串。我想把它放在一个26字符串的字符串中。

如何

我可以填充剩余字符的空间.. ??
是我需要的做一个while循环填充字符串的空间？

不确定你想要什么，但以下程序可能是一个很好的开始。

#include< string.h>

int main（void）
{char small [] =" abc" ;; / *字符串由4个字符组成* /
char large [26] ="" ;; / *由26个字符组成的字符串* /

/ *复制字符串小到大，终止''\ 0''排除* /
memcpy（大，小，sizeof小 - 1）;
/ *将剩余的字符设置为'' - ''* /
memset（大+ sizeof小 - 1，'' - ''，sizeof large - sizeof small）;
/ *用''\ 0''终止大，使其成为一个字符串* /
大[sizeof large - 1] = 0;

/ * large包含现在字符串abc ---------------------- * /

返回0;
}

sorry . i got one more problem
i got a string with 4 char. i want to put that in a string with 26 char. how
can i fill space on the remain char.. ??
is that i need to do a while loop do fill the space for the string?

please help!

thanks
Jack

解决方案

"JC" <[e-mail]> wrote in message news:bl*********@rain.i-cable.com...

i got a string with 4 char. i want to put that in a string with 26 char. how can i fill space on the remain char.. ??
is that i need to do a while loop do fill the space for the string?

Not sure what you want exactly, but the following program might be a good
start.
#include <string.h>

int main(void)
{
char small[] = "abc"; /* string consisting of 4 characters */
char large[26] = ""; /* string consisting of 26 characters */

/* copy the string small into large, the terminating ''\0'' excluded */
memcpy(large, small, sizeof small - 1);
/* set the remaining chararacters in large to ''-'' */
memset(large + sizeof small - 1, ''-'', sizeof large - sizeof small);
/* terminate large with a ''\0'' to make it a string */
large[sizeof large - 1] = 0;

/* large contains now the string "abc----------------------" */

return 0;
}

In article <bl*********@rain.i-cable.com>, JC wrote:

sorry . i got one more problem
i got a string with 4 char. i want to put that in a string with 26 char. how
can i fill space on the remain char.. ??
is that i need to do a while loop do fill the space for the string?

#include <string.h>

/* ... */

char shorty[] = "foo"; /* four chars, including termination */
char lengthy[26] = { 0 };

/* ... */

strcpy(lengthy, shorty);

The string terminator at shorty[3] will be copied over to the
lenthy string, terminating it at lengthy[3]. You don''t need to
"fill out" the string with anything.

If you *do* want to fill the remainder of the string (elements
with index 4 through to 25), then memset() will do this for you:

memset(&(lengthy[4]), ''?'', 21);
lengthy[25] = ''\0'';

A for-loop will do the same thing:

int i;
for (i = 4; i < 25; ++i) {
lengthy[i] = ''?'';
}
lengthy[25] = ''\0'';

The string will still be terminated at index 3 though, unless
you overwrite the ''\0'' at that position with something else.

--
Andreas K?h?ri

thanks.
this is very useful coding.
"dis" <di*@hotmail.com> wrote in message
news:bl**********@news3.tilbu1.nb.home.nl...

"JC" <[e-mail]> wrote in message news:bl*********@rain.i-cable.com...

i got a string with 4 char. i want to put that in a string with 26 char.

how

can i fill space on the remain char.. ??
is that i need to do a while loop do fill the space for the string?

Not sure what you want exactly, but the following program might be a good
start.
#include <string.h>

int main(void)
{
char small[] = "abc"; /* string consisting of 4 characters */
char large[26] = ""; /* string consisting of 26 characters */

/* copy the string small into large, the terminating ''\0'' excluded */
memcpy(large, small, sizeof small - 1);
/* set the remaining chararacters in large to ''-'' */
memset(large + sizeof small - 1, ''-'', sizeof large - sizeof small);
/* terminate large with a ''\0'' to make it a string */
large[sizeof large - 1] = 0;

/* large contains now the string "abc----------------------" */

return 0;
}

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