Use list assignment i.e. for two variables $a, $b: ($a,$b) = ($b,$a)
#include<stdio.h> void main() { int a=2,b=4; printf("Program for swapping two numbers "); printf("Numbers before swapping"); printf("a=%d and b=%d",a,b); a=((a+b)-(b=a)); printf("Numbers after swapping"); printf("a=%d and b=%d",a,b); getch(); }
void swap (int* a, int* b) { if (!a !b) return; // can't swap a pointer to null *a^=*b^=*a^=*b; }
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; }
#include<iostream> void swap(int* x, int* x){ (*x)^=(*y)^=(*x)^=(*y); } int main() { int a, b; a=10; b=20; std::cout<<"Before swap: a="<<a<<", b="<<b<<std::endl; swap(&a,&b); std::cout<<"After swap: a="<<a<<", b="<<b<<std::endl; return(0); }
a=a+b; b=a-b; a=a-b;
To swap two variables without using a third variable, use exclusive or manipulation... a ^= b; b ^= a; a ^= b;
Consider the following declarations:int x = 0;int y = 1;In order to swap the values, we need to use a temporary variable:int t = x;x = y;y = t;However, it is possible to swap the values without using a third variable:x ^= y ^= x ^= y;
It means that you swap the values of that variables EX: -==- before swapping :- Variable1 = 5; variable2 = 10; after swapping :- Variable1 = 10; variable2 = 5;
Use list assignment i.e. for two variables $a, $b: ($a,$b) = ($b,$a)
#include<stdio.h> void main() { int a=2,b=4; printf("Program for swapping two numbers "); printf("Numbers before swapping"); printf("a=%d and b=%d",a,b); a=((a+b)-(b=a)); printf("Numbers after swapping"); printf("a=%d and b=%d",a,b); getch(); }
void swap (int* a, int* b) { if (!a !b) return; // can't swap a pointer to null *a^=*b^=*a^=*b; }
a=a^b; b=a^b; a=a^b;
example: x = x-y; y = y-x; x = y-x; <><><> It is not possible to swap two variables without using a temp variable. The code in the answer above, while clever, does not swap the variables. It will exchange the variable values for certain values of x and y, e.g. when x and y are small integers. If x and y are other values, such as strings, pointers, Infinity, NaN (not a number), floating point, or near the limits of the representation (causing over/underflow) then the code will not "swap" the values. <---> Note: there is absolutely no point in swapping two variables without using temopral variable, it's just a typical homework question, already asked here countless times. Another variation: a ^= b; b ^= a; a ^= b;
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; }
Use a temporary variable. Example (swap variables "a" and "b"): int a = 5; int b = 10; // Swap the variables int temp; temp = a; a = b; b = temp; System.out.println("a = " + a); System.out.println("b = " + b);Use a temporary variable. Example (swap variables "a" and "b"): int a = 5; int b = 10; // Swap the variables int temp; temp = a; a = b; b = temp; System.out.println("a = " + a); System.out.println("b = " + b);Use a temporary variable. Example (swap variables "a" and "b"): int a = 5; int b = 10; // Swap the variables int temp; temp = a; a = b; b = temp; System.out.println("a = " + a); System.out.println("b = " + b);Use a temporary variable. Example (swap variables "a" and "b"): int a = 5; int b = 10; // Swap the variables int temp; temp = a; a = b; b = temp; System.out.println("a = " + a); System.out.println("b = " + b);
#include<iostream> void swap(int* x, int* x){ (*x)^=(*y)^=(*x)^=(*y); } int main() { int a, b; a=10; b=20; std::cout<<"Before swap: a="<<a<<", b="<<b<<std::endl; swap(&a,&b); std::cout<<"After swap: a="<<a<<", b="<<b<<std::endl; return(0); }