删除空格 [英] removing Spaces

问题描述

每当我收到固定长度的字符串时，就说15（未使用的部分

将在收到前填充Spaces），我想从

END中删除空格，直到遇到非空格字符。

假设：testBuf是接收此固定长度字符串的缓冲区。 />

printf（" [％s]"，testBuf）;

输出：

[Hello World]


如何删除空格成为[Hello World]？任何快速有效的算法，如使用strchr等？


谢谢。

解决方案

-----原始消息-----
来自：Magix [mailto：ma *** @ asia.com] 发布到：c
会话：删除空格
主题：删除空格


<每当我收到固定长度的字符串时，就说15（
未使用的部分
将在收到前填充空格），我想从
空格> END，直到遇到非空格字符。
让我们说：testBuf是接收此固定长度字符串的缓冲区。

printf（" [％s]"，testBuf ）;
输出：
[Hello World]

如何删除空格成为[Hello World]？任何使用strchr等快速且高效的算法？

谢谢。

这应该让你开始。但是你需要包含一些

错误检查。


#include< stdio.h>

#include< ; string.h>

#include< ctype.h>


int main（无效）

{

char testBuff [] =" Hello World！ " ;;


int len;


len = strlen（testBuff）-1;


while（isspace（testBuff [len]））

len--;

testBuff [len] =''\ 0'';


printf（" Buffer = [％s] \ n"，testBuff）;


返回0;

}


GST

" Magix" <毫安*** @ asia.com>在消息中写道

news：41 ********** @ news.tm.net.my ...

<每次我收到一个固定长度的字符串，比方说15（未使用的
部分将在收到前填充Spaces），我想从
END中移除空格直到遇到非space char。
让我们说：testBuf是接收这个固定长度字符串的缓冲区。

printf（" [％s]"，testBuf）;
输出：
[Hello World]

如何删除空格成为[Hello World]？任何快速有效的算法，如使用strchr等？

谢谢。

我有这样的东西，但是任何更好的方法？


void str_delete（char * sourc，int pos，int len）

{

int i，j;


--pos;

j = strlen（sourc）;

i = pos + len;

while（i< = j）

{

sourc [pos] = sourc [i];

++ i;

++ pos;

}

}


int isNonSpaceFound = 0;

for（i =（strlen（testBuf）-1）; i> 0; i--）

{

if（testBuf [i]！= 0x20） ）

{

isNonSpaceFound = 1;

}

if（isNonSpaceFound == 0）

{

str_delete（testBuf，strlen（testBuf），1）;

}

}

" Turner，GS（Geoff）"写道：

来自：Magix [mailto：ma *** @ asia.com]

每当我收到固定长度的字符串，比方说15（
未使用的部分将在收到前填充Spaces），我希望
从END中删除空格，直到我遇到非空格的角色。比方说：testBuf是接收这个
<< fix-length string。
printf（" [％s]"，testBuf）;
输出：
[Hello World]

如何删除空格成为[Hello World]？任何快速且高效的C算法，如使用strchr等？

这应该可以让你入门。但是你需要包含一些
错误检查。

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

int main（void）
{char testBuff [] =" Hello World！ " ;;
int len;

len = strlen（testBuff）-1;
** while（isspace（testBuff [len]））
** len - ;
** testBuff [len] =''\ 0'';
printf（" Buffer = [％s] \ n"，testBuff）;
返回0 ;
}

不会那样工作。将标有**的行替换为：


while（len&&（''''== testBuff [len]））

testBuff [len--] =''\ 0'';


-

丘吉尔和布什都可以被视为战时领袖，只是<秘书处和埃德先生都是马匹。
。 - 詹姆斯罗德斯。

我们一直都知道，不顾一切的自我利益是坏的

道德。我们现在知道这是糟糕的经济学 - FDR

Hi,

Everytime I received a fix-length of string, let say 15 (the unused portion
will filled with Spaces before receive), I want to remove the Spaces from
END until I encounter a non-space char.
let say: testBuf is the buffer that receive this fix-length string.

printf("[%s]",testBuf);
Output:
[Hello World ]

How to remove the spaces to become [Hello World] ? Any fast and effcient C
algorithm like using strchr,etc?

Thanks.

解决方案

-----Original Message-----
From: Magix [mailto:ma***@asia.com]
Posted At: 09 September 2004 10:25
Posted To: c
Conversation: removing Spaces
Subject: removing Spaces
Hi,

Everytime I received a fix-length of string, let say 15 (the
unused portion
will filled with Spaces before receive), I want to remove the
Spaces from
END until I encounter a non-space char.
let say: testBuf is the buffer that receive this fix-length string.

printf("[%s]",testBuf);
Output:
[Hello World ]

How to remove the spaces to become [Hello World] ? Any fast
and effcient C
algorithm like using strchr,etc?

Thanks.

This should get you started. But you will need to include some
error checking though.

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

int main (void)
{
char testBuff[]="Hello World! ";

int len ;

len = strlen(testBuff)-1;

while(isspace(testBuff[len]))
len--;
testBuff[len]=''\0'';

printf("Buffer = [%s]\n",testBuff);

return 0;
}

GST

"Magix" <ma***@asia.com> wrote in message
news:41**********@news.tm.net.my...

Hi,

Everytime I received a fix-length of string, let say 15 (the unused portion will filled with Spaces before receive), I want to remove the Spaces from
END until I encounter a non-space char.
let say: testBuf is the buffer that receive this fix-length string.

printf("[%s]",testBuf);
Output:
[Hello World ]

How to remove the spaces to become [Hello World] ? Any fast and effcient C
algorithm like using strchr,etc?

Thanks.

I have something like this, but any better way ?

void str_delete(char * sourc, int pos, int len)
{
int i, j;

--pos;
j=strlen(sourc);
i=pos+len;
while (i <= j)
{
sourc[pos]=sourc[i];
++i;
++pos;
}
}

int isNonSpaceFound=0;
for (i=(strlen(testBuf)-1);i>0;i--)
{
if (testBuf[i]!=0x20)
{
isNonSpaceFound =1;
}
if (isNonSpaceFound==0)
{
str_delete(testBuf,strlen(testBuf),1);
}
}

"Turner, GS (Geoff)" wrote:

From: Magix [mailto:ma***@asia.com]

Everytime I received a fix-length of string, let say 15 (the
unused portion will filled with Spaces before receive), I want
to remove the Spaces from END until I encounter a non-space
char. let say: testBuf is the buffer that receive this << fix-length string.
printf("[%s]",testBuf);
Output:
[Hello World ]

How to remove the spaces to become [Hello World] ? Any fast
and effcient C algorithm like using strchr,etc?

This should get you started. But you will need to include some
error checking though.

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

int main (void)
{
char testBuff[]="Hello World! ";
int len ;

len = strlen(testBuff)-1;
** while(isspace(testBuff[len]))
** len--;
** testBuff[len]=''\0'';
printf("Buffer = [%s]\n",testBuff);
return 0;
}

Won''t work like that. Replace the lines marked with ** with:

while (len && ('' '' == testBuff[len]))
testBuff[len--] = ''\0'';

--
"Churchill and Bush can both be considered wartime leaders, just
as Secretariat and Mr Ed were both horses." - James Rhodes.
"We have always known that heedless self-interest was bad
morals. We now know that it is bad economics" - FDR

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