在ANSI / ISO C ++中操作和写入字节（8位）？ [英] Manipulating and writing bytes (8 bits) in ANSI/ISO C++?
问题描述
我正在写一个简单的图像文件导出器作为学校项目的一部分。要实现我选择的图像格式，我需要使用big-endian
字节，其中''byte''当然意味着''8位''，而不是'' 函数sizeof（char）的 ''。它似乎我可以使用bitset< 8>
了。在我的代码中代表一个字节---如果
你有一个更好的建议，我欢迎它---但这仍然留给我
的问题如何将这些位集以符合标准的方式写入图像文件中作为
big-endian字节。有什么想法吗？
Kristian
2005年4月30日星期六04:49:19 +0300 ，Kristian Nybo
< kr ****** @ begone.spam.iki.fi>写道：
我正在写一个简单的图像文件导出器作为学校项目的一部分。为了实现我选择的图像格式，我需要使用big-endian
字节，其中''byte''当然意味着''8位''，而不是''sizeof（char）'' 。它好像我可以使用bitset< 8>在我的代码中表示一个字节 - 如果你有更好的建议，我欢迎它 - 但这仍然让我想到如何将这些位集写入图像文件的问题
标准兼容的大端字节。任何想法？
Kristian
在我看来，你在考虑不止一个问题。
如果我没有记错，字节顺序与文件，机器和
数据包有关，但与字节本身无关。
如果你想要8位字节从一个不是8位的字节，你可以移动
和/或填充字节。
但是如果你只关心big-endian /小端机器
兼容性，一种解决方案是将文件写入一个字节，时间为
。对于写入的文件，机器将处理机器特定的字节顺序
。如果你正在处理一个文件格式
指定字节顺序，当然你需要知道你从哪个
字节开始写入数据。
Charles
ChasW写道：
在我看来你正在考虑不止一个问题在这里。
如果我没弄错的话，字节顺序与文件，机器和数据包有关，但与字节本身有关。
如果你想要8位字节从一个不是8位的字节，你可以移动
和/或填充字节。
当然，你是对的，但我怎么把它写成8位字节？如果
char是机器上的9位代码正在运行，我该如何
只写8位？
但是如果你只关心big-endian / little endian机器的兼容性，一个解决方案就是在
时间将文件写出一个字节。对于写入的文件，机器将处理机器特定的字节顺序。
文件格式指定字节应该是big-endian，所以我需要
确保它们以这种方式编写，无论代码运行的机器是否使用big-endian或little-endian字节。也就是说，
我不希望机器处理字节序，因为我认为
会导致机器上出现''big-endian files'' big-endian字节
和little-endian文件在一个带有little-endian字节的文件上。你的意思是
别的吗？
干杯，
Kristian
< blockquote> Kristian Nybo写道：
我正在写一个简单的图像文件导出器作为学校项目的一部分。为了实现我选择的图像格式，我需要使用big-endian
字节，其中''byte''当然意味着''8位''，而不是''sizeof（char）'' 。它好像我可以使用bitset< 8>在我的代码中表示一个字节 - 如果你有更好的建议，我欢迎它 - 但这仍然让我想到如何将这些位集写入图像文件的问题
标准兼容的大端字节。任何想法？
Kristian
也许这会有所帮助吗？
man htonl
问候，
拉里
-
反垃圾邮件地址，更改每个' 'X'''''。''直接回复。
Hi,
I''m writing a simple image file exporter as part of a school project. To
implement my image format of choice I need to work with big-endian
bytes, where ''byte'' of course means ''8 bits'', not ''sizeof(char)''. It
seems that I could use bitset<8> to represent a byte in my code --- if
you have a better suggestion, I welcome it --- but that still leaves me
with the question of how to write those bitsets to an image file as
big-endian bytes in a standard-compliant manner. Any ideas?
Kristian
On Sat, 30 Apr 2005 04:49:19 +0300, Kristian Nybo
<kr******@begone.spam.iki.fi> wrote:
Hi,
I''m writing a simple image file exporter as part of a school project. To
implement my image format of choice I need to work with big-endian
bytes, where ''byte'' of course means ''8 bits'', not ''sizeof(char)''. It
seems that I could use bitset<8> to represent a byte in my code --- if
you have a better suggestion, I welcome it --- but that still leaves me
with the question of how to write those bitsets to an image file as
big-endian bytes in a standard-compliant manner. Any ideas?
Kristian
Seems to me you are contemplating more than one issue here.
If I am not mistaken, endianness pertains to files, machines, and
packets, but not the bytes themselves.
If you want 8 bit bytes from a byte that is not 8 bits, you can shift
and / or pad the byte.
But if you are just concerned with big-endian / little endian machine
compatibility, one solution is to write the file out one byte at a
time. The machine will handle machine specific endianness with regard
to the file written. If you are dealing with a file format that
specifies byte order, of course you will need to be aware of which
byte you start writing the data from.
Charles
ChasW wrote:
Seems to me you are contemplating more than one issue here.
If I am not mistaken, endianness pertains to files, machines, and
packets, but not the bytes themselves.
If you want 8 bit bytes from a byte that is not 8 bits, you can shift
and / or pad the byte.
You''re right, of course, but how would I write it as an 8 bit byte? If
char is, say, 9 bits on the machine the code is running on, how do I
write only 8 bits?
But if you are just concerned with big-endian / little endian machine
compatibility, one solution is to write the file out one byte at a
time. The machine will handle machine specific endianness with regard
to the file written.
The file format specifies that the bytes should be big-endian, so I need
to make sure they''re written that way regardless of whether the machine
the code is running on uses big-endian or little-endian bytes. That is,
I don''t want the machine to handle the endianness because I assume that
would result in ''big-endian files'' on a machine with big-endian bytes
and little-endian files on one with little-endian bytes. Did you mean
something else?
Cheers,
Kristian
Kristian Nybo wrote:Hi,
I''m writing a simple image file exporter as part of a school project. To
implement my image format of choice I need to work with big-endian
bytes, where ''byte'' of course means ''8 bits'', not ''sizeof(char)''. It
seems that I could use bitset<8> to represent a byte in my code --- if
you have a better suggestion, I welcome it --- but that still leaves me
with the question of how to write those bitsets to an image file as
big-endian bytes in a standard-compliant manner. Any ideas?
Kristian
Perhaps this would help?
man htonl
Regards,
Larry
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