std::copy 原始数据使用 ostream_iterator 以十六进制格式复制到 cout<uint8_t> 打印出未格式化的数据。 为什么？

Question

The idea is to have a vector containing arbitrary binary data and outputting the bytes it contains to stdout in hexadecimal notation. I use std::copy to copy the bytes from the input vector to stdout. The problem is that the code prints out 0 followed by raw binary data. Here is the code:

auto vec = std::vector<uint8_t> {'h', 'e', 'l', 'l', 'o'};
std::cout << std::hex << std::setfill('0') << std::setw(2);
std::copy(vec.begin(), vec.end(), std::ostream_iterator<uint8_t>(std::cout));

The application prints out "0hello". I would expect it to print out "68656C6C6F". What can be the case?

Answer 1

Simple change, use int instead of uint8_t :

std::copy(vec.begin(), vec.end(), std::ostream_iterator<int>(std::cout));

If you want it in uppercase, use std::uppercase .

operator<< has non-member overloads for char , signed char , and unsigned char . See: uint8_t iostream behavior .

Answer 2

Simply replace std::cout in

std::ostream_iterator<uint8_t>(std::cout)

with std::cout<<std::hex and uint8_t with int

std::ostream_iterator<int>(std::cout<<std::hex)

This works because << operator returns reference to the modified stream, which is passed further to the iterator. Also you can combine it with other modificators such as std::showbase , eg

std::ostream_iterator<int>(std::cout<<std::showbase<<std::hex, " ")