The only way to swap two values using call by value semantics is to pass pointer variables by value. A pointer is a variable that stores an address. Passing a pointer by value copies the address, the value of the pointer, not the pointer itself. By passing the addresses of the two values to be swapped, you are effectively passing those values by reference. Both C and C++ use pass by value semantics by default, however C++ also has a reference data type to support native pass by reference semantics. By contrast, Java uses pass by reference semantics by default.
In C, to swap two variables using pass by value:
void swap (int* p, int* q) {
int t = *p;
*p = *q;
*q = t;
}
In C++, to swap two variables using pass by reference:
void swap (int& p, int& q) {
std::swap (p, q);
}
Note that C++ is more efficient because std::swap uses move semantics; there is no temporary variable required to move variables. With copy semantics, a temporary is required. However, with primitive data types, there is a way to swap values without using a temporary, using a chain of exclusive-or assignments:
void swap (int* p, int* q) {
*p^=*q^=*p^=*q;
}
To swap two numbers N1 and N2, using a third variable T... T = N1; N1 = N2; N2 = T;
How do you do. I am doing well thank you. Swap two number by using reference operators A tough assignment, it will make you think. I think you are confusing reference operators with pointers. Were I you I would study the section on pointers in your text book or course material.
flow chart to swap two number
constant pointer and character pointer
By dereferencing the pointer variable. This can be achieved in two ways: typedef struct s { int i; float f; }; void f (struct s* p) { int x = p->i; /* using pointer to member operator */ float y = (*p).f; /* using dereference operator */ } The two methods are functionally equivalent.
void swap (int &pa, int &pb) { *pa ^= *pb; *pb ^= *pa; *pa ^= *pb; }
To swap two numbers N1 and N2, using a third variable T... T = N1; N1 = N2; N2 = T;
To swap two variables without using a third variable, use exclusive or manipulation... a ^= b; b ^= a; a ^= b;
swap (int *a, int *b) { *a ^= *b; *b ^= *a; *a ^= *b; }
a += b; b -= a; a -= b;
You cannot swap two numbers using call by value, because the called function does not have access to the original copy of the numbers.Swap with call by reference... This routine uses exclusive or swap without temporary variable.void swap (int *a, int *b) {*a ^= *b;*b ^= *a;*a ^= *b;return;}
How do you do. I am doing well thank you. Swap two number by using reference operators A tough assignment, it will make you think. I think you are confusing reference operators with pointers. Were I you I would study the section on pointers in your text book or course material.
void main() { int a=2,b=5; b=a+b; a=b-a; b=b-a; getch(); }
Not necessarily. If you swap them, the difference will be negative.
flow chart to swap two number
They had None. They had two songs that reached number 2: Fire and Slow Hand
constant pointer and character pointer