将十六进制数据输出到文件的正确方法是什么？ [英] What is the correct way to output hex data to a file?

问题描述

我已阅读过 [ostream]<< hex<< 0x（十六进制值），但我有一些问题

I've read about [ostream] << hex << 0x[hex value], but I have some questions about it

（1）我定义了我的文件流，输出，使用 output.open（BWhite.bmp，ios :: binary）; 是否会使输出<< 操作中的 hex 参数多余？

(1) I defined my file stream, output, to be a hex output file stream, using output.open("BWhite.bmp",ios::binary);, since I did that, does that make the hex parameter in the output<< operation redundant?

（2）
如果我有一个整数值，我想存储在文件中，我使用这个：

(2) If I have an integer value I wanted to store in the file, and I used this:

int i = 0;
output << i;

会存储在小端序还是大端序？将根据程序在哪个计算机上执行或编译来更改内容？

would i be stored in little endian or big endian? Will the endi-ness change based on which computer the program is executed or compiled on?

此值的大小取决于运行的计算机吗？我需要使用十六进制参数吗？

Does the size of this value depend on the computer it's run on? Would I need to use the hex parameter?

（3）有没有办法输出原始十六进制数字到文件？如果我想要该文件的十六进制数字43，我应该使用什么？

(3) Is there a way to output raw hex digits to a file? If I want the file to have the hex digit 43, what should I use?

输出<< 0x43 和输出<< hex<< 0x43 都输出ASCII 4，然后输出ASCII 3。

output << 0x43 and output << hex << 0x43 both output ASCII 4, then ASCII 3.

输出这些十六进制数字的目的是为.bmp文件创建标题。

The purpose of outputting these hex digits is to make the header for a .bmp file.

推荐答案

格式化的输出运算符<< ：格式化输出。

The formatted output operator << is for just that: formatted output. It's for strings.

因此， std :: hex 流操纵器指示流输出数字作为字符串格式如十六进制。

As such, the std::hex stream manipulator tells streams to output numbers as strings formatted as hex.

如果要输出原始二进制数据，请仅使用未格式化的输出函数 basic_ostream :: put 和 basic_ostream :: write 。

If you want to output raw binary data, use the unformatted output functions only, e.g. basic_ostream::put and basic_ostream::write.

你可以输出一个类似这样的int：

You could output an int like this:

int n = 42;
output.write(&n, sizeof(int));

此输出的字节序将取决于架构。如果你想有更多的控制，我建议如下：

The endianness of this output will depend on the architecture. If you wish to have more control, I suggest the following:

int32_t n = 42;
char data[4];
data[0] = static_cast<char>(n & 0xFF);
data[1] = static_cast<char>((n >> 8) & 0xFF);
data[2] = static_cast<char>((n >> 16) & 0xFF);
data[3] = static_cast<char>((n >> 24) & 0xFF);
output.write(data, 4);

此示例将输出一个32位整数作为小尾数，无论平台的字节顺序如何。但是，如果 char 已签名，请小心转换。

This sample will output a 32 bit integer as little-endian regardless of the endianness of the platform. Be careful converting that back if char is signed, though.

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