Variable initialization do not eval last expression

[VincentDamour]

Fix for this issue

With this PR, variable initialization will not be push into the expressions array.
Inline variable initialization with an output is still valid. let x = 1; x will output 1

[StephenGrider]

@funny-vince Thanks for the contribution, this change would be a solid improvement.

One small issue: while the statement

const number = 1; 
number;

Does print '1', it appears that the nearly equivalent expression

const numbers = [1,2,3]; 
numbers;

Will not print anything, even though

const numbers = [1,2,3]; numbers;

Does give the expected result.

The 'eval' function tends to return some unpredictable results at times, it may be worth confirming that this behavior isn't being caused by eval.

[VincentDamour]

@StephenGrider Hi, thanks for your answer. Fixing the issue you have mentioned will break the evaluation of loop with variable initialization in their declaration. For now I cannot help you, I'm sorry for this. Your code works great with eval(), but I found many small bugs while trying to fix this issue. I don't know what are you plan and your goal for JSPlaygrounds, but if you want a neat playground (like JS Bin and JSFiddle) there is probably many things to think about and a lot of code to change/write. If you need help with this, I will be more than happy to help you, but for now, I cannot fix this issue without changing a lot of the existing code.

[StephenGrider]

no problem