写入4位用的ofstream二进制文件 [英] Writing 4 bits to a binary file with ofstream

问题描述

如果我有0之间的无符号整数 - 16，我想将其写入到一个二进制文件，而不整个字节写入，如何将一个移位位来实现它。

0-16意味着我只需要4位，所以我应该能够存储在一个字节右2不同的数字？

以下code写入1号1个字节：

 的std :: ofstream的文件;
file.open（TEST.bin，烧写的std :: IOS ::二进制|性病:: IOS ::出来）;
焦炭oneByteNum =（char）的fourByteNum; //一些4字节的整数转换为1字节字符
file.write（安培; oneByteNum，sizeof的（炭））;
 

使用bitshifts，我如何能实现2个数字为1字节？
我想读一些出字节将是一个相似的，反两个步骤呢？

解决方案

 字符oneByteWithTwoNums =（oneByteNum1＆LT;＆LT; 4）+（oneByteNum2＆安培;为0x0F）;
 

If I have an unsigned integer between 0 - 16 and I want to write it to a binary file without writing a whole byte to it, how would one shift bits to achieve it?

0-16 means I only need 4 bits, so I should be able to store 2 different numbers in a single byte right?

The following code writes 1 number to 1 byte:

std::ofstream file;
file.open("test.bin", std::ios::binary|std::ios::out);
char oneByteNum = (char)fourByteNum; // Converting from some 4 byte integer to a 1 byte char
file.write(&oneByteNum ,sizeof(char));

Using bitshifts, how can I achieve 2 numbers in 1 byte? I imagine reading the number out of the byte would be a similar, inverse 2 step process too?

解决方案

char oneByteWithTwoNums = (oneByteNum1 << 4) + (oneByteNum2 & 0x0f);

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