使用十六进制表示法将缓冲区转换为字符串流： [英] Getting a buffer into a stringstream in hex representation:

问题描述

如果我有一个缓冲区：

  uint8_t buffer [32] 
  

它完全填充了值，我怎么能得到它到字符串流，十六进制表示，

  std :: stringstream ss; 
 for（int i = 0; i <32; ++ i）
 {
 ss< std :: hex<< buffer [i]; 
} 
  

但是当我从stringstream中取出字符串时，具有值<

例如，如果数组中的字节1和2都是{32} {4}，那么我们只需要一个字符来表示。 } my stringstream会有：

  204而不是2004 
  

我可以对stringstream应用格式化以添加0填充吗？我知道我可以用sprintf这样做，但是这些流已经用于很多信息，这将是一个很大的帮助，以这种方式实现。

解决方案

  std :: stringstream ss; 
 ss<< std :: hex<< std :: setfill（'0'）; 
 for（int i = 0; i <32; ++ i）
 {
 ss< std :: setw（2）<< static_cast< unsigned>（buffer [i]）; 
} 
  

If I had a buffer like:

uint8_t buffer[32];

and it was filled up completely with values, how could I get it into a stringstream, in hexadecimal representation, with 0-padding on small values?

I tried:

std::stringstream ss;
for (int i = 0; i < 32; ++i)
{
    ss << std::hex << buffer[i];
}

but when I take the string out of the stringstream, I have an issue: bytes with values < 16 only take one character to represent, and I'd like them to be 0 padded.

For example if bytes 1 and 2 in the array were {32} {4} my stringstream would have:

204 instead of 2004

Can I apply formatting to the stringstream to add the 0-padding somehow? I know I can do this with sprintf, but the streams already being used for a lot of information and it would be a great help to achieve this somehow.

解决方案

std::stringstream ss;
ss << std::hex << std::setfill('0');
for (int i = 0; i < 32; ++i)
{
    ss << std::setw(2) << static_cast<unsigned>(buffer[i]);
}

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