C ++ iostream>>运算符的行为与get（）unsigned char不同 [英] C++ iostream >> operator behaves differently than get() unsigned char

问题描述

我正在处理一段代码来做一些压缩，我写了一个比特流类。

我的比特流类跟踪我们正在读取的当前位和当前字节（unsigned char）。

我注意到，从文件读取下一个无符号字符是不同的，如果我使用>>运算符vs get

我只是想知道为什么我得到不同的结果。

ex：

  this-> m_inputFileStream.open（inputFile，std :: ifstream :: binary）; 
 unsigned char currentByte; 
 this-> m_inputFileStream>>> currentByte; 
  

vs。

 code> this-> m_inputFileStream.open（inputFile，std :: ifstream :: binary）; 
 unsigned char currentByte; 
 this-> m_inputFileStream.get（（char&）currentByte）; 
  

其他信息：

我读的字节是0x0A但是当使用>>它会读它为0x6F

我不知道他们甚至是相关的？ （它们不是彼此的2的补码？）

>>运算符也被定义为对无符号字符有效（见 c ++ istream类参考

operator>> 或运算符< < / / code>。你会得到很难跟踪的奇怪的bug，它们对单元测试也是有弹性的，除非你知道要查找什么。读取一个uint8实例将在9失败。

编辑：

  #include< iostream> 
 #include< sstream> 
 #include< cstdint> 
 
 void test（char r）{
 std :: cout< < r<< std :: endl; 
 char t ='！'; 
 std :: ostringstream os（std :: ios :: binary）; 
 os< ; r; 
 if（！os.good（））std :: cout<<os not good< std :: endl; 
 std :: istringstream（os.str（），std :: ios :: binary）; 
是>> t; 
 if（！is.good（））std :: cout<< 不好< std :: endl; 
 std :: cout<< std :: hex<< （uint16_t）r 
<< vs< std :: hex<< （uint16_t）t < std :: endl; 
} 
 
 int main（int argc，char ** argv）{
 test（'z'）; 
 test（'\\\
'）; 
 return 0; 
} 
  

产生：

 测试z 
 7a vs 7a 
测试
 
不好
a vs 21 
  

我想这从来不会是先验明显的。

I was working on a piece of code to do some compression, and I wrote a bitstream class.

My bitstream class kept track of the current bit we are reading and the current byte (unsigned char).

I noticed that reading the next unsigned character from the file was done differently if I used the >> operator vs get() method in the istream class.

I was just curious why I was getting different results?

ex:

this->m_inputFileStream.open(inputFile, std::ifstream::binary);   
unsigned char currentByte;
this->m_inputFileStream >> currentByte;

vs.

this->m_inputFileStream.open(inputFile, std::ifstream::binary);
unsigned char currentByte;
this->m_inputFileStream.get((char&)currentByte);

Additional Info:

To be specific the byte I was reading was 0x0A however when using >> it would read it as 0x6F

I'm not sure how they're even related ? (they're not the 2s complement of each other?)

The >> operator is also defined to work for unsigned char as well however (see c++ istream class reference

解决方案

If you aren't parsing text, don't use operator>> or operator<<. You'll get weird bugs that are hard to track down. They are also resilient to unit tests, unless you know what to look for. Reading a uint8 for instance will fail on 9 for instance.

edit:

#include <iostream>
#include <sstream>
#include <cstdint>

void test(char r) {
        std::cout << "testing " << r << std::endl;
        char t = '!';
        std::ostringstream os(std::ios::binary);
        os << r;
        if (!os.good()) std::cout << "os not good" << std::endl;
        std::istringstream is(os.str(), std::ios::binary);
        is >> t;
        if (!is.good()) std::cout << "is not good" << std::endl;
        std::cout << std::hex << (uint16_t)r 
             << " vs " << std::hex << (uint16_t)t << std::endl;
}

int main(int argc, char ** argv) {
        test('z');
        test('\n');
        return 0;
}

produces:

testing z
7a vs 7a
testing 

is not good
a vs 21

I suppose that would never have been evident a priori.

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