比较char [2]和short [英] Compare char[2] with short
问题描述
我需要一种在内存块中搜索char [2]数组的方法
" DA"使用指向短的指针。理想情况下我想写
类似于：
short * data = ...一些数据......;
int j = 0;
while（data [j]！= *（（short *）" DA"））j ++;
但这不是不行。 char [2]显然有一个等价的16位
I need a way to search through a block of memory for a char[2] array
"DA" using a pointer to a short. Ideally I would like to write
something like:
short *data = ... some data...;
int j = 0;
while( data[j] != *((short*) "DA") ) j++;
But this doesn''t work. The char[2] obviously has an equivalent 16-bit
value so how do I get that info in a simple way?
推荐答案
spasmous写道：
spasmous wrote:
我需要一种方法来搜索内存块以获取char [2]数组
" ; DA"使用指向短的指针。理想情况下我想写
类似于：
short * data = ...一些数据......;
int j = 0;
while（data [j]！= *（（short *）" DA"））j ++;
但这不是不行。 char [2]显然具有相当的16位
I need a way to search through a block of memory for a char[2] array
"DA" using a pointer to a short. Ideally I would like to write
something like:
short *data = ... some data...;
int j = 0;
while( data[j] != *((short*) "DA") ) j++;
But this doesn''t work. The char[2] obviously has an equivalent 16-bit
value so how do I get that info in a simple way?
只需按字符比较。
while（data [j]！=''D''&& ; data [j + 1]！=''A''））j ++;
别忘了添加数组末尾的检查！
-
Ian Collins。
Just compare character by character.
while( data[j] != ''D'' && data[j+1] != ''A'') ) j++;
Don''t forget to add checking for the end of the array!
--
Ian Collins.
5月17日，1：13 * pm，Ian Collins< ian -n ... @ hotmail.comwrote：
On May 17, 1:13*pm, Ian Collins <ian-n...@hotmail.comwrote:
spasmouswrote：
spasmouswrote:
我需要一种方法来搜索通过char [2]数组的内存块
" DA"使用指向短的指针。理想情况下我想写
类似于：
I need a way to search through a block of memory for a char[2] array
"DA" using a pointer to a short. Ideally I would like to write
something like:
short * data = ... some data ...;
int j = 0;
while（data [j]！= *（（short *）" DA"））j ++;
short *data = ... some data...;
int j = 0;
while( data[j] != *((short*) "DA") ) j++;
但这不起作用。 char [2]显然具有相当的16位
But this doesn''t work. The char[2] obviously has an equivalent 16-bit
value so how do I get that info in a simple way?
只需按字符比较。
while（data [j]！=''D''&& ; data [j + 1]！=''A''））j ++;
别忘了添加数组末尾的检查！
Just compare character by character.
while( data[j] != ''D'' && data[j+1] != ''A'') ) j++;
Don''t forget to add checking for the end of the array!
也许我混淆了，但至少在我的平台上char是8位而
short是16位。因此，一个数据[j]与两个字符相同。
Maybe I''m mixed up, but at least on my platform char is 8-bit and
short is 16-bit. So one data[j] is the same as two chars.
在comp.lang.c中，spasmous写道：
In comp.lang.c, spasmous wrote:
我需要一种方法来搜索内存块以获取char [2]数组
DA使用指向短的指针。理想情况下我想写
类似于：
short * data = ...一些数据......;
int j = 0;
while（data [j]！= *（（short *）" DA"））j ++;
但这不是不行。 char [2]显然具有相当的16位
I need a way to search through a block of memory for a char[2] array
"DA" using a pointer to a short. Ideally I would like to write
something like:
short *data = ... some data...;
int j = 0;
while( data[j] != *((short*) "DA") ) j++;
But this doesn''t work. The char[2] obviously has an equivalent 16-bit
value so how do I get that info in a simple way?
具体实施[1]（但仍然是IIRC，标准C中的合法）方式
将是
while（data [j]！=''DA''）++ j;
[1] ISO C 9899-1999 6.4.4.4。第10点（语义）说包含多个字符的
整数字符常量的值......是实现定义的
。 ''DA''是一个常数。
-
Lew Pitcher
Master Code Code& JOAT-in-training |已注册的Linux用户＃112576
http://pitcher.digitalfreehold.ca/ |可根据要求提供GPG公钥
---------- Slackware - 因为我知道我在做什么。 ------
An implementation specific[1] (but still, IIRC, legal in standard C) way
would be to
while (data[j] != ''DA'') ++j;
[1] ISO C 9899-1999 6.4.4.4. Point 10 (Semantics) says "The value of an
integer character constant containing more than one character ... is
implementation-defined." ''DA'' is such a constant.
--
Lew Pitcher
Master Codewright & JOAT-in-training | Registered Linux User #112576
http://pitcher.digitalfreehold.ca/ | GPG public key available by request
---------- Slackware - Because I know what I''m doing. ------
这篇关于比较char [2]和short的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
