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Home/ Questions/Q 8165695
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Editorial Team
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Editorial Team
Asked: June 6, 20262026-06-06T19:37:10+00:00 2026-06-06T19:37:10+00:00

I have a very simple arithmetic operator but am at my wits end why

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I have a very simple arithmetic operator but am at my wits end why it doesn’t return 2. The code below returns 1. I thought that x++ equates to x = x + 1;

CODE

var x = 1;
document.write(x++);

However if I run the code as follows, it returns 2 as expected

CODE

var x = 1;
document.write(++x);

What am I doing wrong?

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  1. Editorial Team
    Editorial Team
    2026-06-06T19:37:11+00:00Added an answer on June 6, 2026 at 7:37 pm

    PostIncrement(variable++) & PostDecrement(variable–)

    When you use the ++ or -- operator after the variable, the variable’s value is not incremented/decremented until after the expression is evaluated and the original value is returned. For example x++ translates to something similar to the following:

    document.write(x);
    x += 1;
    

    PreIncrement(++variable) & PreDecrement(–variable)

    When you use the ++ or -- operator prior to the variable, the variable’s value is incremented/decremented before the expression is evaluated and the new value is returned. For example ++x translates to something similar to the following:

    x += 1;
    document.write(x);
    

    The postincrement and preincrement operators are available in C, C++, C#, Java, javascript, php, and I am sure there are others languages. According to why-doesnt-ruby-support-i-or-i-increment-decrement-operators, Ruby does not have these operators.

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