Question

How to swap two numbers without using third variable in JavaScript

Answer 1

let a = 5, b = 13;
      
a = a + b;
b = a - b;
a = a - b;   
 
console.log("a = " + a);
console.log("b = " + b);

 
 
/*
run:
  
a = 13
b = 5
   
*/

 

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Answer 2

let a = 5, b = 13;
      
a = a + b - (b = a);    
 
console.log("a = " + a);
console.log("b = " + b);

 
 
/*
run:
  
a = 13
b = 5
   
*/

 

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Answer 3

let a:number = 7, b = 13;
      
a = a + b, b = a - b, a = a - b;
 
console.log("a = " + a);
console.log("b = " + b);

 
 
/*
run:
  
"a = 13" 
"b = 7" 
   
*/

 

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Answer 4

let a = 5, b = 13;
      
a = a ^ b;
b = a ^ b;
a = b ^ a; 
 
console.log("a = " + a);
console.log("b = " + b);

 
 
/*
run:
  
a = 13
b = 5
   
*/

 

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Answer 5

let a = 5, b = 13;
      
a = b - ~a - 1;
b = a + ~b + 1;
a = a + ~b + 1;
 
console.log("a = " + a);
console.log("b = " + b);

 
 
/*
run:
  
a = 13
b = 5
   
*/

 

70+ SQL courses for beginners and professionals

Answer 6

let a = 5, b = 13;
      
[a, b] = [b, a];
 
console.log("a = " + a);
console.log("b = " + b);

 
 
/*
run:
  
a = 13
b = 5
   
*/

 

70+ SQL courses for beginners and professionals