References can not be re initialized. They always refer to first referent.
Reference arithmetic is not allowed. Operation effects on actual referent.
#include<iostream>
using namespace std;
int main()
{
int i=10;
int& j = i;
int & k =j;
cout<<"i: "<<i<<endl;
cout<<"j: "<<j<<endl;
cout<<"k: "<<k<<endl;
cout<<"&i: "<<&i<<endl;
cout<<"&j: "<<&j<<endl;
cout<<"&k: "<<&k<<endl;
//reference once initialized can not be reinitialed. point to first referent only.
int m=10;
int n=20;
int &x=m;
cout<<"x="<<x<<endl;//10
x=n;//this is not reassignment. actually its modifying value of m using x. this can be verified taking address of m and x as both would be same.
cout<<"x="<<x<<endl;//20
cout<<"m="<<m<<endl;//m gets modified to 20 now
cout<<"&x="<<&x<<endl;//0x7fff9671caf0
cout<<"&m="<<&m<<endl;//0x7fff9671caf0
//reference arithmatic is not allowed
int a=10;
int b=20;
int& ar=a;
cout<<"a="<<a<<endl;//10
cout<<"ar="<<ar<<endl;//10
ar++;//referant gets incremented not reference
cout<<"a="<<a<<endl;//11
cout<<"ar="<<ar<<endl;//11
return 0;
}
/*
i: 10
j: 10
k: 10
&i: 0x7fffee78603c
&j: 0x7fffee78603c
&k: 0x7fffee78603c
x=10
x=20
m=20
&x=0x7fffee786040
&m=0x7fffee786040
a=10
ar=10
a=11
ar=11
*/
Showing posts with label reference. Show all posts
Showing posts with label reference. Show all posts
C++ reference FAQ interview questions
//1.References are nothing but constant pointer.
//2.So,references once initialized can not refer to other.
//2.1 Reference to reference not allowed.No error but. it will start pointing to the first variable
//3. reference arithmatic not allowed.No error in reference arithmatic but changes reflected in referant not in reference.
//4.Reference to array allowed.//int (&r)[5];
//5.Array of reference not allowed because we can not find the address of array. ultimately refer to first variable.
//6.Unlike pointers,refernce are automatically gets dereferenced.No need to use *(value at address).
//7.Unlike pointers, reference arithmatic like increment will increase the referant's value and not the address
//8.Reference are used to pass by reference to get the changes reflected in calling function.
//9.Pass by reference avoids copy of large value when structure like data are passed as arguments.
//10. Its unsafe to return local variable by value because variable goes out of scope once conrol returns out of the function defintion.
//Solution is us local static variable or dynamic memory allocation so that scope of the variable will through out the program.
#include<iostream>
using namespace std;
int gv;
int main()
{
//reference nothing but constant pointer means once initialized can not be reinitialized.
int s=10;
int *const h=&s;//s is const pointer so can not hold another address
//h=&m;//error: assignment of read-only variable ‘h’
int i=10,m=2;
int &j=i;// j is reference to a
//reference to refernce not allowed. No error but points to the first variable
int &k=j;//reference to reference no error but//point 2.1
k=20;
cout<<i<<endl;//20
//pointer arithmatic are not allowed. no error but changes gets reflected in first variable
k++;//point 7
cout<<i<<endl;//21
//referece can not be reinitialized but direct values can be given
int a=1;
k=1;
cout<<k<<endl;
//pass by reference to get the values reflected in calling function
int x=10,y=20;
swap(x,y);
cout<<x<<" " <<y<<endl;//20,10
//if return type is reference then function call can be made in left side
int &f();
f()=5;
cout<<"gv="<<gv<<endl;//5//instead of returned reference 100, f() is assigned to 5
//problem of local ovariable returning by value
int d;
int ff(d);
cout<<"d="<<d<<endl;//Garbage, because return value went out of scope in definition itself.
return 0;
}
void swap(int x, int y)
{
int t;
t=x;
x=y;
y=x;
}
//return by reference
int &f()
{
gv=100;
return gv;
}
//problem of returning local variable by value
int ff(int t)
{
int& lv=t;
lv=lv+44;
return lv;//returning local variable by value.
}//scope of lv over here
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//2.So,references once initialized can not refer to other.
//2.1 Reference to reference not allowed.No error but. it will start pointing to the first variable
//3. reference arithmatic not allowed.No error in reference arithmatic but changes reflected in referant not in reference.
//4.Reference to array allowed.//int (&r)[5];
//5.Array of reference not allowed because we can not find the address of array. ultimately refer to first variable.
//6.Unlike pointers,refernce are automatically gets dereferenced.No need to use *(value at address).
//7.Unlike pointers, reference arithmatic like increment will increase the referant's value and not the address
//8.Reference are used to pass by reference to get the changes reflected in calling function.
//9.Pass by reference avoids copy of large value when structure like data are passed as arguments.
//10. Its unsafe to return local variable by value because variable goes out of scope once conrol returns out of the function defintion.
//Solution is us local static variable or dynamic memory allocation so that scope of the variable will through out the program.
#include<iostream>
using namespace std;
int gv;
int main()
{
//reference nothing but constant pointer means once initialized can not be reinitialized.
int s=10;
int *const h=&s;//s is const pointer so can not hold another address
//h=&m;//error: assignment of read-only variable ‘h’
int i=10,m=2;
int &j=i;// j is reference to a
//reference to refernce not allowed. No error but points to the first variable
int &k=j;//reference to reference no error but//point 2.1
k=20;
cout<<i<<endl;//20
//pointer arithmatic are not allowed. no error but changes gets reflected in first variable
k++;//point 7
cout<<i<<endl;//21
//referece can not be reinitialized but direct values can be given
int a=1;
k=1;
cout<<k<<endl;
//pass by reference to get the values reflected in calling function
int x=10,y=20;
swap(x,y);
cout<<x<<" " <<y<<endl;//20,10
//if return type is reference then function call can be made in left side
int &f();
f()=5;
cout<<"gv="<<gv<<endl;//5//instead of returned reference 100, f() is assigned to 5
//problem of local ovariable returning by value
int d;
int ff(d);
cout<<"d="<<d<<endl;//Garbage, because return value went out of scope in definition itself.
return 0;
}
void swap(int x, int y)
{
int t;
t=x;
x=y;
y=x;
}
//return by reference
int &f()
{
gv=100;
return gv;
}
//problem of returning local variable by value
int ff(int t)
{
int& lv=t;
lv=lv+44;
return lv;//returning local variable by value.
}//scope of lv over here
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