使用sprintf将十六进制0xAABBCC转换为字符串“ AA：BB：CC”

Question

I have a hex number 0xaabbcc which I would like convert and format into a char string as AA:BB:CC .

using sprintf(myStr, %X, 0xaabbcc); results in char myStr[] = "AABBCC";

Is it possible to use sprintf or some other function to convert and format 0xaabbcc to AA:BB:CC ?

Answer 1

You need to surround the specifier string with quotes. The printf family receives a const char* as the format string. %x alone outside of a string means something modulo by x, and will result in compiler error if there's no integer before % like in your case

After that just split the bytes you want to print

unsigned v = 0xAABBCCU;
sprintf(myStr, "%02X:%02X:%02X", v >> 16, (v >> 8) & 0xFFU, v & 0xFFU); // or
sprintf(myStr, "%02X:%02X:%02X", v >> 16, (uint8_t)(v >> 8), (uint8_t)v);

However since you're using C++ it'll be safer to use std::string

std::stringstream myStr;
myStr << std::hex << std::setfill('0')
      << std::setw(2) << (v >> 16)
      << std::setw(2) << ((v >> 8) & 0xFFU)
      << std::setw(2) << (v & 0xFFU)

Answer 2

If you want to stick with printf and you are using GCC (or maybe Clang, but I can't find anything documented) then you can use printf cusomization .

Frankly @phuclv's answer is probably better.