fread / fwrite可移植性问题 [英] fread/fwrite Portability Issues
问题描述
-----开始PGP签名消息-----
哈希：SHA1
嘿所有。
我正试图找到一种方法（便携）将32位整数值写入
文件。就目前而言，我一直在使用这样的东西：
#include< stdio.h>
/ *写入val到f指向的文件。 * /
int write32（unsigned long val，FILE * f）
{
unsigned char c;
int i;
/ *确保指针有效。 * /
if（！f）
{
fprintf（stderr，NULL file pointer.\ n）;
返回1;
}
/ *将4个字节从LSB写入MSB。 * /
for（i = 0; i< 4; i ++）
{
c =（val>（i * 8） ）& 0xff;
if（fwrite（& c，1,1，f）！= 1）
{
/ *确保写作发生了。 * /
fprintf（stderr，Write error.\ n）;
返回1;
}
}
/ *正常退出。 * /
返回0;
}
这似乎有用，但我被告知char并不总是8位。这个
意味着在具有较大字符大小的big-endian系统上，我将写入
全部为零。
有人可以推荐替代品吗？
- -
问候，
Jonathan Lamothe
/ *
*糟糕。内核试图访问一些糟糕的页面。我们将不得不以极端的偏见终止事情。
* /
die_if_kernel（" Oops"， regs，error_code）;
- 来自linux / arch / i386 / mm / fault.c
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版本：GnuPG v1.4.2（GNU / Linux）
评论：将GnuPG与Thunderbird一起使用 - http://enigmail.mozdev.org
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= RAEB
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-----BEGIN PGP SIGNED MESSAGE-----
Hash: SHA1
Hey all.
I''m trying to find a way to (portably) write 32-bit integer values to a
file. As it stands, I''ve been using something like this:
#include <stdio.h>
/* Write val to the file pointed to by f. */
int write32(unsigned long val, FILE *f)
{
unsigned char c;
int i;
/* Make sure the pointer is valid. */
if(!f)
{
fprintf(stderr, "NULL file pointer.\n");
return 1;
}
/* Write the 4 bytes from LSB to MSB. */
for(i = 0; i < 4; i++)
{
c = (val >(i * 8)) & 0xff;
if(fwrite(&c, 1, 1, f) != 1)
{
/* Ensure the writing occurred. */
fprintf(stderr, "Write error.\n");
return 1;
}
}
/* Exit normally. */
return 0;
}
This seems to work, but I''m told that char is not always 8 bits. This
means that on big-endian systems with larger char sizes, I''ll be writing
all zeros to the file.
Can anyone suggest an alternative?
- --
Regards,
Jonathan Lamothe
/*
* Oops. The kernel tried to access some bad page. We''ll have to
* terminate things with extreme prejudice.
*/
die_if_kernel("Oops", regs, error_code);
-- From linux/arch/i386/mm/fault.c
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Comment: Using GnuPG with Thunderbird - http://enigmail.mozdev.org
iD8DBQFEwm7wq9nD47x87JYRAoMGAJ957wTuvop8ijiHMOrvaT 81c6b6+wCfQulN
DiVY3WI5ORK+oK2YJgF0sas=
=RAEB
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推荐答案
Jonathan Lamothe写道：
Jonathan Lamothe wrote:
-----开始PGP签名消息-----
哈希：SHA1
嘿所有。
我正试图找到一种方法（便携）写32位整数值到一个
文件。就目前而言，我一直在使用这样的东西：
-----BEGIN PGP SIGNED MESSAGE-----
Hash: SHA1
Hey all.
I''m trying to find a way to (portably) write 32-bit integer values to a
file. As it stands, I''ve been using something like this:
< snip>
<snip>
c =（val>（i * 8））& 0xff;
if（fwrite（& c，1,1，f）！= 1）
c = (val >(i * 8)) & 0xff;
if(fwrite(&c, 1, 1, f) != 1)
否，转换为缓冲区然后调用fwrite一次。你正在调用单个字节的
函数，这非常低效。
No, convert to a buffer then call fwrite once. You''re calling the
function for single bytes and that''s terribly inefficient.
这似乎有用，但我'告诉我，char并不总是8位。这个
意味着在具有较大字符大小的big-endian系统上，我将写入
全部为该文件的零。
This seems to work, but I''m told that char is not always 8 bits. This
means that on big-endian systems with larger char sizes, I''ll be writing
all zeros to the file.
嗯，没有。如果您将数据存储在较低的32位中，那么总是可以使用
工作。也就是说，无论是否有结尾
无符号x = 0x1234;
x>> = 8;
x现在将等于0x12，无论平台是大，小甚至是
中端。
Tom
Um, no. If you store data in the lower 32-bits then that will always
work. That is, regardless of endianess
unsigned x = 0x1234;
x >>= 8;
x will now equal 0x12 regardless if the platform is big, little or even
middle endian.
Tom
Tom St Denis写道：
Tom St Denis wrote:
Jonathan Lamothe写道：
Jonathan Lamothe wrote:
- ---开始PGP签名消息-----
哈希：SHA1
嘿所有。
我正试图找到一种方法（便携地）将32位整数值写入
文件。就目前而言，我一直在使用这样的东西：
-----BEGIN PGP SIGNED MESSAGE-----
Hash: SHA1
Hey all.
I''m trying to find a way to (portably) write 32-bit integer values to a
file. As it stands, I''ve been using something like this:
< snip>
<snip>
c =（val>（i * 8））& 0xff;
if（fwrite（& c，1,1，f）！= 1）
c = (val >(i * 8)) & 0xff;
if(fwrite(&c, 1, 1, f) != 1)
否，转换为缓冲区然后调用fwrite一次。你在单个字节上调用
函数，这非常低效。
No, convert to a buffer then call fwrite once. You''re calling the
function for single bytes and that''s terribly inefficient.
为什么效率低下？无论如何，fwrite可能会将字节放在
a缓冲区中。
Why would it be inefficient ? fwrite will probably put the bytes in
a buffer anyway.
这似乎有用，但我'告诉我，char并不总是8位。这个
意味着在具有较大字符大小的big-endian系统上，我将写入
全部为该文件的零。
This seems to work, but I''m told that char is not always 8 bits. This
means that on big-endian systems with larger char sizes, I''ll be writing
all zeros to the file.
嗯，没有。如果您将数据存储在较低的32位中，那么总是可以使用
工作。也就是说，无论endianess
Um, no. If you store data in the lower 32-bits then that will always
work. That is, regardless of endianess
确实，该程序是正确的。我无法想象为什么Jonathan认为
如果char大于8位，零将被存储到文件中。
Spiros Bousbouras
spi ... @ gmail.com写道：
spi...@gmail.com wrote:
确实，程序是正确的。我无法想象为什么乔纳森认为如果char大于8位，零将被存储到文件中。
Indeed , the programme is correct. I can''t imagine why Jonathan thinks
that zeros will be stored to the file if char is greater than 8 bits.
他可能在64位平台上认为
无符号x = 0x1234;
意味着
x = 0x1234000000000000
和8的移位得到LSB零。
当然，[对于OP]在内存中它可以存储为
12 34 00 00 00 00 ...
但是当你将它用作数据类型时，操作已经明确定义了
行为。所以x& 255"在你修改x之前总是为0x34。
你是对的，你不能只记忆或写下未签名的[或
无论什么]类型到一个文件，并期望代码在其他地方工作。
但是，如果你[正确]掩盖了不变的字节，它将作为
所需。
Tom
He''s probably thinking that on a 64-bit platform that
unsigned x = 0x1234;
would mean
x = 0x1234000000000000
And the shift by 8 gets the LSB zeroes.
Sure, [for the OP] in memory it may be stored as
12 34 00 00 00 00 ...
But when you use it as a data type the operations have well defined
behaviours. So "x & 255" is always 0x34 until you modify x.
You''re correct that you can''t just memcpy or fwrite the unsigned [or
whatever] types to a file and expect the code to work elsewhere.
However, if you [correctly] mask off the invidual bytes it will work as
desired.
Tom
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