我想让一个类足够灵活,以处理基于模板中的 typename 类型的显式转换。怎么做到呢?我猜班级将不得不处理转换
请注意,这似乎需要大量代码,但我尽可能地缩短了它。
这是我想做的。
template<typename T> void fooAndBar(T x)
{
uint64_t count = 0;
//i want next line to work for x being either uint64_t or bigNumber
while( x <= (T)1000 ) {
count++;//do something
}
}
int main() {
uint64_t y = 1;
fooAndBar(y);
bigNumber z;
fooAndBar(z);
return 0;
}
bigNumber 定义为:
#include <vector>
typedef unsigned long long int uill;
class bigNumber //just a class that can handle reaaaly big numbers very striped down
{
public:
bigNumber();
~bigNumber() {}
bool operator<=(bigNumber rightSide);
template<typename T> static bigNumber convertToBigNumber(T in);
private:
std::vector<uill> digits;
};
bigNumber::bigNumber() { digits.push_back(1); }
bool bigNumber::operator<=(bigNumber rightSide) {
// dummy function
return digits.back() <= rightSide.digits.back();
}
template<typename T> static bigNumber bigNumber::convertToBigNumber(T in) {
bigNumber bn;
bn.digits.push_back(in);
return bn;
}
但我被困在单独的功能上。一个用于 uint64_t,一个用于 bigNumber,如下所示
void foo(uint64_t x) {
uint64_t count = 0;
// NB x is an uint64_t
while( x <= 1000 ) {
count++;//do something
}
}
void bar(bigNumber x) {
uint64_t count = 0;
// NB x is a bigNumber
while( x <= bigNumber::convertToBigNumber(1000) ) {
count++;//do something
}
}
int main() {
uint64_t y=1;
foo(y);
bigNumber z ;
bar(z);
return 0;
}
提前感谢您的帮助。这个问题需要更好的标签,任何建议都会有所帮助。