问题
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.
回答1:
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2:
Look at the stream modifiers: std::setw
and std::fill
.
来源:https://stackoverflow.com/questions/7639656/getting-a-buffer-into-a-stringstream-in-hex-representation