numeric_limits::is_specialized
Tests if a type has an explicit specialization defined in the template class numeric_limits.
Syntax
static const bool is_specialized = false;
Return Value
true if the type has an explicit specialization defined in the template class; false if not.
Remarks
All scalar types other than pointers have an explicit specialization defined for template class numeric_limits.
Example
// numeric_limits_is_specialized.cpp
// compile with: /EHsc
#include <iostream>
#include <limits>
using namespace std;
int main( )
{
cout << "Whether float objects have an explicit "
<< "specialization in the class: "
<< numeric_limits<float>::is_specialized
<< endl;
cout << "Whether float* objects have an explicit "
<< "specialization in the class: "
<< numeric_limits<float*>::is_specialized
<< endl;
cout << "Whether int objects have an explicit "
<< "specialization in the class: "
<< numeric_limits<int>::is_specialized
<< endl;
cout << "Whether int* objects have an explicit "
<< "specialization in the class: "
<< numeric_limits<int*>::is_specialized
<< endl;
}
Whether float objects have an explicit specialization in the class: 1 Whether float* objects have an explicit specialization in the class: 0 Whether int objects have an explicit specialization in the class: 1 Whether int* objects have an explicit specialization in the class: 0
Requirements
Header: <limits>
Namespace: std