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void fun(int *b)
cout<<"value of pointer in fun()"<<"\n"<<b;
cout<<"the value of c is"<<*b;
cout<<"previous value of a is "<<*p;
cout<<"value of pointer in main :"<<"\n "<<p;
cout<<"after calling fun ,new value of a is "<<*p;
how is the o/p remaining the same in main() even after we change the pointer's to point to another variable in func()??
The parameter is passed by value, so this statement: 'b=&c' changes only the copied value, not the original.
PS: you really should use printf, it is much more flexible, eg:
printf ("main, beforore 'fun': a=%d, p=%p\n", a, p);
thanx.i thnk i get it..but jst to make sure when i change the argument to be passed as refernce, the o/p changes..so does this actually mean when we pass a reference to a variable,another copy of that reference is not created in func()??
void fun(int &b)
cout<<"previous value of a is "<<p;
cout<<"after calling fun ,new value of a is "<<p;