C++ 以十六进制表示将缓冲区放入字符串流中：
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Getting a buffer into a stringstream in hex representation:
提问by John Humphreys - w00te
If I had a buffer like:
如果我有一个缓冲区，如：
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?
并且它完全被值填满了，我怎么能把它变成一个字符串流，以十六进制表示，在小值上填充 0？
I tried:
我试过：
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.
但是当我从字符串流中取出字符串时，我遇到了一个问题：值 < 16 的字节只需要一个字符来表示，我希望它们被填充为 0。
For example if bytes 1 and 2 in the array were {32} {4} my stringstream would have:
例如，如果数组中的字节 1 和 2 是 {32} {4}，我的 stringstream 将具有：
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.
我可以将格式应用于 stringstream 以某种方式添加 0-padding 吗？我知道我可以用 sprintf 来做到这一点，但是流已经被用于大量信息，以某种方式实现这一点会很有帮助。
回答by Dean Povey
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]);
}
回答by Thomas Matthews
Look at the stream modifiers: std::setwand std::fill.
查看流修饰符： std::setw和std::fill。
