我有一个uint128_t
类,将其值存储为uint64_t UPPER, LOWER;
,我不知道如何重载operator<<
,以便在我传入std::cout
时,值将以十进制正确打印。目前,我正在做
std::ostream & operator<<(std::ostream & stream, uint128_t const & rhs){
if (rhs.upper()) // if the upper value has a non-zero digit
stream << rhs.upper();
// i need some way to pad this so that the number of 0s between
// upper and lower is correct
stream << rhs.lower();
return stream;
我该怎么办?
编辑:
示例:
如果uint128_t变量有UPPER = 1
和LOWER = 1
,我希望该流包含小数值(1 << 64) + 1
答案 0 :(得分:1)
假设您已正确实施了除法运算符和模数运算符,则可以执行以下操作:
std::ostream & operator<<(std::ostream & stream, uint128_t const & rhs){
if(rhs.upper() == 0)
return stream << rhs.lower();
char buffer[50];
char *cp = buffer + 49;
*cp = 0;
while(rhs > 0)
{
--cp;
*cp = (rhs % 10) + '0';
rhs /= 10;
}
return stream << cp;
}