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A write program determines a pointer variable is not
Supplements:
- C ++ still supports variable parameter function in C language.
- C ++ compiler matches the call priority
- Overloaded functions.
- Function template.
- Variable argument function.
Ideas:
- The variables are divided into two types: pointer and non-pointer.
- Writing functions:
- Returns true if the pointer variable called.
- Return false non-pointer variables call.
Chemical reaction with template function becomes a function parameter:
Programming experiment: determination Pointer
#include <iostream>
#include <string>
using namespace std;
class Test
{
public:
Test()
{
}
virtual ~Test()
{
}
};
template
<typename T>
char IsPtr(T* v) // match pointer
{
return 'd';
}
int IsPtr(...) // match non-pointer
{
return 0;
}
int main(int argc, char *argv[])
{
int i = 0;
int* p = &i;
cout << "p is a pointer: " << IsPtr(p) << endl; // true
cout << "i is a pointer: " << IsPtr(i) << endl; // false
Test t;
Test* pt = &t;
cout << "pt is a pointer: " << IsPtr(pt) << endl; // true
cout << "t is a pointer: " << IsPtr(t) << endl; // false
return 0;
}
Appeals achieve defect versions exist:
- Variadic function object parameter can not be resolved, it could cause the program to crash!
A further challenge:
- How to let the compiler exactly matching function, but not the actual call? sizeof.
#include <iostream>
#include <string>
using namespace std;
class Test
{
public:
Test()
{
}
virtual ~Test()
{
}
};
template
<typename T>
char IsPtr(T* v) // match pointer
{
return 'd';
}
int IsPtr(...) // match non-pointer
{
return 0;
}
#define ISPTR(p) (sizeof(IsPtr(p)) == sizeof(char))
int main(int argc, char *argv[])
{
int i = 0;
int* p = &i;
cout << "p is a pointer: " << ISPTR(p) << endl; // true
cout << "i is a pointer: " << ISPTR(i) << endl; // false
Test t;
Test* pt = &t;
cout << "pt is a pointer: " << ISPTR(pt) << endl; // true
cout << "t is a pointer: " << ISPTR(t) << endl; // false
return 0;
}
References: