Is it guaranteed that the primary template will be selected when a class template with full specialization of long is materialized with int, as shown in the following code?
#include<iostream>
template<typename T>
structure A
{
A( )
{
std::cout<<"primary"<<std::endl;
}
};
template<>
structure A<long>
{
A( )
{
std::cout<<"specialized"<<std::endl;
}
};
int main()
{
A<int>b2;
}
output
primary
I've been implementing it in the hope that the output will look like the one above, but when I thought about it carefully, I didn't know if this was a defined behavior.
14.7 Template installation and specialization
or
Template-Related
for cppreference.com
I read as much as I could, but I did my best!
Thank you for your cooperation.
Quotations are acceptable, and articles, chapter numbers, etc. JIS X3014:2003
Perhaps you are confused that you can call In the case of Try commenting on some of the specializations or specifying the values of the arguments in suffixes to see how they work.func(long) (where the argument type is int instead of long) especially for the general integer promotion 4.5, but implicitly for this integer promotion template type resolution, the candidate function selection logic per 13.3.1, 13.3.2 and 13.3.3 is considered as a priority over the general integer promotion.#include<iostream>
template <typename T> structure A {
A(T) {std::cout<<"primary"<<std::endl;}
};
template<>struct A<long>{
A(long) {std::cout<<"long"<<std::endl;}
};
template<>struct A<short>{
A(short) {std::cout<<"short"<<std::endl;}
};
template<>struct A<char>{
A(char) {std::cout<<"char"<<std::endl;}
};
template<>struct A<int>{
A(int) {std::cout<<"int"<<std::endl;}
};
int main()
{
Aa1(0);
Aa2('\0');
Aa3(0L);
Aa4(0U);
}
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