问题
Considering following struct and template function, does each use of the function with a different value for "num" build a new instance of the function or since const numbers<num> & nums parameter is a reference and would be implemented as a pointer all uses with different values for "num" can be directed to one instance of the function?
template<size_t num>
struct numbers{
public:
unsigned int nums[num];
};
template<size_t num>
void print(const numbers<num> & nums,size_t size){
for (int i=0;i<size;i++)
cout <<nums.nums[i]<<' ';
cout <<'\n';
}
回答1:
As far as you (the programmer) are concerned, print<2>() and print<3>() are separate functions. Whilst the compiler may theoretically be able to make this optimisation, it probably won't happen in pratice. The best way to find out for sure is to look at the generated assembler.
回答2:
Yes, compiler will generate function for every case of num.
回答3:
Yes, each instantiation of the template with a new set of parameter values yelds a new function.
来源：https://stackoverflow.com/questions/4133307/here-different-instances-are-generated-for-this-template-function-or-not