Why doesn't the shorthand arithmetic operator ++ after the variable name return 2 in the following statement?

Question

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;

C

Answer 1

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.