便携式ascii-hex转换 [英] portable ascii-hex conversion
问题描述
全部，
我有一系列字符需要转换为整数值。
每个字符依次从函数中读取' 'nextch''和十六进制数字是由isxdigit函数识别的
- 所以我在看''0'' - ''9''，''A'' - ' 'Z''
和'''' - ''z''。
这就是我所拥有的：
int num = 0;
int ch = nextch（）; / * nextc获得下一个字符值* /
while（isxdigit（ch））
{
if（isdigit（ ch））
ch = ch - ''0''; / *这是便携式我相信* /
其他
ch =（ch& ~0x20） - ''A''+ 10; / *不确定这是否正常* /
num = num * 0x10 + ch;
ch = nextch（）;
}
如果你看看while（）循环中的if-else语句，你会看到
我如何尝试转换''ch ''从字符值到数字值在
范围内0-15（含）。但我怀疑（（ch& ~0x20） - ''A''+
10）表达式：
它假设'' '' - ''F''是连续的值
它假设'''' - ''''''是连续的，并且总是比它们的价格高出$
''大写''对应物。
这些假设是否正确？我猜这段代码是不便携的，所以
有没有人有一个整洁的（呃）建议？
p.s.我从lcc编译器源代码中派生出这个代码...
James
All,
I have a series of characters which I need to convert to integer values.
Each character is read in turn from a function ''nextch'', and hex-digits are
identified by the isxdigit function - so I''m looking at ''0'' - ''9'', ''A'' - ''Z''
and ''a'' - ''z''.
Here is what I''ve got:
int num = 0;
int ch = nextch(); /* nextc obtains the next character value */
while(isxdigit(ch))
{
if(isdigit(ch))
ch = ch - ''0''; /* this is portable I believe */
else
ch = (ch & ~0x20) - ''A'' + 10; /* not sure if this is ok */
num = num * 0x10 + ch;
ch = nextch();
}
If you look at the if-else statement inside the while() loop, you will see
how I attempt to convert ''ch'' from a character-value to a numeric value in
the range 0-15 inclusive. But I have doubts about the ((ch & ~0x20) - ''A'' +
10) expression:
It assumes that ''A'' - ''F'' are consecutive values
It assumes that ''a'' - ''f'' are consecutive, and are always 0x20 above their
''uppercase'' counterparts.
Are these assumptions correct? I''m guessing the code is non-portable, so
does anyone have a neat(er) suggestion?
p.s. I derived this code from the lcc compiler sourcecode...
James
推荐答案
在文章< lt********************@pipex.net>，
James Brown< no*@home.netwrote：
In article <lt********************@pipex.net>,
James Brown <no*@home.netwrote:
>我有一系列字符需要转换为整数值。
每个字符依次从函数''nextch''读取，并且十六进制-digits由isxdigit函数识别 - 所以我看''0'' - ''9''，'''' - '''Z''
和''一个'' - ''z''。
>I have a series of characters which I need to convert to integer values.
Each character is read in turn from a function ''nextch'', and hex-digits are
identified by the isxdigit function - so I''m looking at ''0'' - ''9'', ''A'' - ''Z''
and ''a'' - ''z''.
>这是我得到的：
>Here is what I''ve got:
> int num = 0;
int ch = nextch（）; / * nextc获取下一个字符值* /
>int num = 0;
int ch = nextch(); /* nextc obtains the next character value */
> while（isxdigit（ch））
>while(isxdigit(ch))
如果是EOF怎么办？
What if it was EOF ?
> {
if（isdigit（ch））
ch = ch - ''0''; / *这是便携式我相信* /
>{
if(isdigit(ch))
ch = ch - ''0''; /* this is portable I believe */
是的。
Yes.
else
ch =（ch& ~0x20） - ''A''+ 10; / *不确定这是否正常* /
else
ch = (ch & ~0x20) - ''A'' + 10; /* not sure if this is ok */
& ~0x20是一个隐藏的toupper（），不能移植到非ASCII。
正如你所想的那样，A到F并不能保证是
结果甚至增加订单。
The & ~0x20 is a hidden toupper() and not portable to non-ASCII.
And as you had thought, ''A'' through ''F'' are not guaranteed to be
consequative or even increasing order.
num = num * 0x10 + ch;
num = num * 0x10 + ch;
如果你的int溢出怎么办？
What if you overflow your int ?
ch = nextch（）;
}
ch = nextch();
}
>这些假设是否正确？我猜测代码是不可移植的，所以
有没有人有一个整洁的（呃）建议？
>Are these assumptions correct? I''m guessing the code is non-portable, so
does anyone have a neat(er) suggestion?
只要isxdigit（ch）获取更多ch，你就没有比你可以处理更多的字符
更多的字符和存储将它们放入缓冲区。
然后strtoul（）指定基数16.
如果您有特殊原因自己处理角色，
然后创建一个大小为UCHAR_MAX的翻译表，
并初始化它，tr [''0''+ i] = i for i从0到9，并且
tr [''A''] = 10，tr [''B''] = 11等，tr [''a''] = 10，tr [''b''] = 11等。 ，
然后进行转换，只需确定isxdigit（ch），如果是这样，那么转换后的值为tr [ch]。是的，这有可能浪费UCHAR_MAX - 26个插槽，但它也是一个便携式单步
转换，没有数学（除了正常的数组索引）
-
编程是在你忙于制定其他计划时发生的事情。
fetch more ch as long as isxdigit(ch) and you haven''t gotten
more chars than you can handle, and store them into a buffer.
Then strtoul() specifying base 16.
If you have particular reasons for handling the characters yourself,
then create a translation table of size UCHAR_MAX,
and initialize it, tr[''0''+i] = i for i from 0 to 9, and
tr[''A''] = 10, tr[''B''] = 11, etc., tr[''a''] = 10, tr[''b''] = 11, etc.,
then to do the conversion, just determine isxdigit(ch) and if so
then the converted value is tr[ch]. Yes, this has the potential
to waste UCHAR_MAX - 26 slots, but it is also a portable single-step
conversion with no math (other than normal array indexing)
--
Programming is what happens while you''re busy making other plans.
" Walter Roberson" ; < ro ****** @ ibd.nrc-cnrc.gc.cawrote in message
news：ei ********** @ canopus.cc.umanitoba.ca ...
"Walter Roberson" <ro******@ibd.nrc-cnrc.gc.cawrote in message
news:ei**********@canopus.cc.umanitoba.ca...
在文章< lt ******************** @ pipex.net>中，
James Brown< no*@home.netwrote：
In article <lt********************@pipex.net>,
James Brown <no*@home.netwrote:
>>我有一系列我需要转换的字符到整数值。
每个字符依次从函数''nextch''读取，而十六进制数字
由isxdigit函数识别 - 所以我在看' '0'' - ''9''，''A'' -
''Z''
和'''' - ''z''。
>>I have a series of characters which I need to convert to integer values.
Each character is read in turn from a function ''nextch'', and hex-digits
are
identified by the isxdigit function - so I''m looking at ''0'' - ''9'', ''A'' -
''Z''
and ''a'' - ''z''.
>>这是我得到的：
>>Here is what I''ve got:
>> int num = 0;
int ch = nextch（）; / * nextc获取下一个字符值* /
>>int num = 0;
int ch = nextch(); /* nextc obtains the next character value */
>> while（isxdigit（ch））
>>while(isxdigit(ch))
如果是EOF怎么办？
What if it was EOF ?
我以为会没事的？ ch将是EOF，这将导致isxdigit
返回（0），并且循环将爆发。这不是会发生什么吗？
I thought it would be ok? ch would be EOF, which would cause isxdigit to
return(0), and the loop would break out. Is this not what would happen?
>
>
>> {
if（isdigit（ch））
ch = ch - ''0''; / *这是便携式我相信* /
>>{
if(isdigit(ch))
ch = ch - ''0''; /* this is portable I believe */
是的。
Yes.
>否则
ch =（ch& ~0x20） - ''A''+ 10; / *不确定这是否正常* /
> else
ch = (ch & ~0x20) - ''A'' + 10; /* not sure if this is ok */
& ~0x20是一个隐藏的toupper（），不能移植到非ASCII。
正如你所想的那样，A到F并不能保证是
结果甚至增加订单。
The & ~0x20 is a hidden toupper() and not portable to non-ASCII.
And as you had thought, ''A'' through ''F'' are not guaranteed to be
consequative or even increasing order.
> num = num * 0x10 + ch;
> num = num * 0x10 + ch;
如果溢出int怎么办？
What if you overflow your int ?
是的，我没有检查溢出，这是我的下一个任务。
yes, I hadn''t gotten as far as checking for overflow, that''s my next task.
>
>
> ch = nextch（）;
}
> ch = nextch();
}
>>这些假设是否正确？我猜测代码是不可移植的，所以
有没有人有一个整洁的（呃）建议？
>>Are these assumptions correct? I''m guessing the code is non-portable, so
does anyone have a neat(er) suggestion?
只要isxdigit（ch）获取更多ch，你就没有比你可以处理更多的字符
更多的字符和存储将它们放入缓冲区。
然后strtoul（）指定基数16.
fetch more ch as long as isxdigit(ch) and you haven''t gotten
more chars than you can handle, and store them into a buffer.
Then strtoul() specifying base 16.
绝对是一个不错的解决方案，但我认为这很难检测溢出？
我的strtoul编译器文档说它在
溢出时返回ULONG_MAX，但是当我遇到
$时如何区分它？ b $ b实际ULONG_MAX值？这就是为什么我对这个东西进行手工编码，这样我就可以在发生这种情况时发出适当的警告信息。
Definitely a nice solution, but I think it will be hard to detect overflows?
My compiler documentation for strtoul says that it returns ULONG_MAX on
overflow, but how do I distinguish this from the case when I encounter the
actual ULONG_MAX value? This is why I am hand-coding this thing, so that I
can emit appropriate warning messages when such things happen.
>
如果您有特殊原因自己处理角色，
然后创建一个大小为UCHAR_MAX的翻译表，
并初始化它，tr [ ''0''+ i] = i for i从0到9，
tr [''A''] = 10，tr [''B''] = 11等。 ，tr [''a''] = 10，tr [''b''] = 11等等，然后
进行转换，只需确定isxdigit（ch），如果是， br />
然后转换后的值是tr [ch]。是的，这有可能浪费UCHAR_MAX - 26个插槽，但它也是一个便携式单步
转换，没有数学（除了正常的数组索引）
-
编程是在你忙于制定其他计划时发生的事情。
>
If you have particular reasons for handling the characters yourself,
then create a translation table of size UCHAR_MAX,
and initialize it, tr[''0''+i] = i for i from 0 to 9, and
tr[''A''] = 10, tr[''B''] = 11, etc., tr[''a''] = 10, tr[''b''] = 11, etc.,
then to do the conversion, just determine isxdigit(ch) and if so
then the converted value is tr[ch]. Yes, this has the potential
to waste UCHAR_MAX - 26 slots, but it is also a portable single-step
conversion with no math (other than normal array indexing)
--
Programming is what happens while you''re busy making other plans.
我绝对会认为这是一个解决方案 - 我希望有一个1/2衬里
（调用一个c-runtime func是理想的），但它看起来像一个查找表
可能是最合适的前进方式。我不太关心
的性能 - 我更喜欢一个简单的循环。
谢谢，
James
I''ll definitely consider this as a solution - I was hoping for a 1/2 liner
(calling a c-runtime func would be ideal), but it looks like a lookup table
may be the most appropriate way forward. I''m not too concerned with
performance though - I would prefer a simple loop above all else.
thanks,
James
>>
>>
>> num = num * 0x10 + ch;
>> num = num * 0x10 + ch;
如果你的int溢出怎么办？
What if you overflow your int ?
是的，我没有检查溢出，这是我的下一个任务。
yes, I hadn''t gotten as far as checking for overflow, that''s my next task.
>>
>>
>> ch = nextch（）;
}
>> ch = nextch();
}
>>>这些假设是否正确？我猜测代码是不可移植的，所以
有没有人有一个整洁的（呃）建议？
>>>Are these assumptions correct? I''m guessing the code is non-portable, so
does anyone have a neat(er) suggestion?
只要isxdigit（ch）获取更多ch，你就没有比你能处理的更多的字符，并将它们存储到缓冲区中。
然后strtoul（）指定基数16.
fetch more ch as long as isxdigit(ch) and you haven''t gotten
more chars than you can handle, and store them into a buffer.
Then strtoul() specifying base 16.
绝对是一个很好的解决方案，但我认为很难检测到
溢出？我对strtoul的编译器文档表示它在溢出时返回
ULONG_MAX，但是我如何区分这与
遇到实际ULONG_MAX值时的情况？这就是为什么我手动编码这个
的东西，以便在发生这样的事情时我可以发出适当的警告信息。
Definitely a nice solution, but I think it will be hard to detect
overflows? My compiler documentation for strtoul says that it returns
ULONG_MAX on overflow, but how do I distinguish this from the case when I
encounter the actual ULONG_MAX value? This is why I am hand-coding this
thing, so that I can emit appropriate warning messages when such things
happen.
ok，所以我读了其余的strtoul文档，并说''errno''设置为
溢出/下溢。看起来这是我的首选解决方案，感谢
帮助。
James
ok, so I read the rest of the strtoul docs and it says ''errno'' is set for
overflow/underflow. Looks like this is my preferred solution, thanks for the
help.
James
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