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
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;
}
#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
如何
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
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;
}
#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
how
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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