how does below code give result as 10
var myVar = 2;
myVar = ++myVar + myVar++ + myVar--;
console.log(myVar);and when directly given as below without assigning to myVar give result as 3 in javascript?
var myVar = 2;
++myVar + myVar++ + myVar--;
console.log(myVar);There are two main factors we need to analyze:
Pre-Increment - The variable is first incremented, and then its value is used:
++x
Post-Increment - The variable value is first used, and then the value is incremented:
x++
With the first example:
var myVar = 2;
myVar = ++myVar + myVar++ + myVar--;
console.log(myVar);
The ++myVar increments the value from 2 to 3 first, so the value 3 is used. Then, for myVar++, the value 3 is used first, and then it is incremented to 4. For myVar--, the 4 is used, and then the value decrements from 4 to 3.
In the end, the expression will be: 3 + 3 + 4 = 10
For the second example:
var myVar = 2;
++myVar + myVar++ + myVar--;
console.log(myVar);
The individual +'s between the myVar are doing nothing. They are adding them together, but there is no variable storing that value. So the only thing that line does is apply the changes of ++myVar, myVar++, and myVar-- to myVar, giving it a value of 3 (from 2 to 3 to 4 to 3).
I hope this helped! Please let me know if you need any further clarification or details :)
As @Teemu said. Reading the documentation is really helpful. Documentation
Maybe this helps also to understand this.
var myVar = 2;
myVar += myVar++;
console.log(myVar);
myVar = 3 + 3 + 4; // 10, so we assign the result of adding all the operands (those being++myVar,myVar++, andmyVar--). The second one evaluates to3 + 3 + 4, but that does nothing with the calculation, so all we see is the side effects of running++and--onmyVar, where it ends up as3.