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Type: Posts; User: ivanandzorro

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  1. Replies
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    So this will work too: x = parseFloat(x);

    So this will work too:


    x = parseFloat(x);
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    Thanks, yes this makes sense. My reading on...

    Thanks, yes this makes sense. My reading on Javascript claims that the interpreter can "figure out" what the data type of a variable should be. I guess it's not perfect especially since the addition...
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    parseInt unexpected behavior

    I am very new to Javascript so please bear with me. I would like to do something simple like pass a variable, have the variable rounded off and then printed. Here' the code I am trying:


    var x...
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