c++有关uint8_t打印异常那些事

描述

代码里用到一个uint8的枚举量，无法通过log打印观测
即使用static_cast<uint8_t>转换依然无法正常输出

原因

看到图片第一反应是ASCII码问题，查询资料进一步印证猜想

uint8_t is an alias for unsigned char, and the iostreams have special overloads for chars that print out the characters rather than formatting numbers.

枚举量范围0-3，而小于32的char对应ASCII不可见符号

解决

原因明确，只要将uint8_t的值还原成对应字符即可正常打印，参考so添加unsigned()解决

uint8_t aa=5;

cout << "value is " << unsigned(aa) << endl;

对于枚举量+aa并不能解决问题

参考

c++ - uint8_t iostream behavior - Stack Overflow
c++ - uint8_t can't be printed with cout - Stack Overflow