Материал из C\C++ эксперт
Use a reference parameter.
#include <iostream>
using namespace std;
void neg(int &i); // i now a reference
int main()
{
int x;
x = 10;
cout << x << " negated is ";
neg(x); // no longer need the & operator
cout << x << "\n";
return 0;
}
void neg(int &i)
{
i = -i; // i is now a reference, don"t need *
}Use reference type eliminates the more confusing pointer notation.
#include <iostream>
using namespace std;
struct stStudent {
char pszName[66],
pszAddress[66],
pszCity[26],
pszState[3],
pszPhone[13];
int icourses;
float GPA;
};
void vByValueCall (stStudent stAStudent);
void vByVariableCall (stStudent *pstAStudent);
void vByReferenceCall (stStudent &rstAStudent);
int main(void)
{
stStudent astLargeClass[100];
astLargeClass[0].icourses = 10;
vByValueCall( astLargeClass[0]);
cout << astLargeClass[0].icourses << "\n"; // icourses still 10
vByVariableCall(&astLargeClass[0]);
cout << astLargeClass[0].icourses << "\n"; // icourses = 20
vByReferenceCall( astLargeClass[0]);
cout << astLargeClass[0].icourses << "\n"; // icourses = 30
}
void vByValueCall(stStudent stAStudent)
{
stAStudent.icourses += 10; // normal structure syntax
}
void vByVariableCall(stStudent *pstAStudent)
{
pstAStudent->icourses += 10; // pointer syntax
}
void vByReferenceCall(stStudent &rstAStudent)
{
rstAStudent.icourses += 10; // simplified reference syntax
}uses reference parameters to swap the values of the variables it is called with
#include <iostream>
using namespace std;
void swap(int &i, int &j);
int main()
{
int a, b, c, d;
a = 1;
b = 2;
c = 3;
d = 4;
cout << "a and b: " << a << " " << b << "\n";
swap(a, b); // no & operator needed
cout << "a and b: " << a << " " << b << "\n";
cout << "c and d: " << c << " " << d << "\n";
swap(c, d);
cout << "c and d: " << c << " " << d << "\n";
return 0;
}
void swap(int &i, int &j)
{
int t;
t = i; // no * operator needed
i = j;
j = t;
}
