为什么std :: stoul会转换负数？
如 http://ideone.com/RdINqa 中所示， std :: stoul 不会为负数抛出 std :: out_of_range ，而是将它们包装起来。为什么是这样？好像-4超出了 unsigned long 类型的范围，因此它应该抛出。
As shown in http://ideone.com/RdINqa, std::stoul does not throw a std::out_of_range for negative numbers, but wraps them around. Why is this? Seems like -4 is out of the range of the type unsigned long so it should throw.
21.5 数值转换
无符号长stoul（const string& str，size_t * idx = 0，int base =
10）;
效果： ... call [s] strtoul（str.c_str（），ptr，base） ...返回转换后的结果（如果有）。
Effects: ...call[s] strtoul(str.c_str(), ptr, base) ... returns the converted result, if any.
抛出： ... out_of_range 如果转换后的值超出返回类型可表示值的范围。
Throws: ...out_of_range if the converted value is outside the range of representable values for the return type.
转换后的价值是 strtoul 返回的值。当然，它的类型为 unsigned long ，因此对于 stoul的返回类型，不能超出可表示值的范围，它也是 unsigned long 。
"Converted value" here is the value returned by strtoul. Which, of course, is of type unsigned long and so can't be outside of the range of representable values for the return type of stoul, which is also unsigned long.
据我所知，只有 stoi 可以抛出 out_of_range ，因为它返回 int 但使用 strtol 会返回 long 。
As far as I can tell, only stoi can throw out_of_range, because it returns int but uses strtol which returns long.
此外，采用C标准指定 strtoul ，则需要接受字符串-4 并返回等于-（无符号长）4 。我不知道为什么要这样指定。
Further, the way C standard specifies strtoul, it is required to accept the string "-4" and return a value equal to -(unsigned long)4. Why it's specified this way, I don't know.