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No reference error when the variable is not an argument #154

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@galpeter

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@galpeter

Consider the following code:

try {
    xxxx;
} catch(e) {
    print(e);
}

Will not print out the expected ReferenceError string. But if we change the code in a way that the xxxx variable is part of an argument list then the exception is correctly raised.

So using the following code:

try {
    print(xxxx);
} catch(e) {
    print(e);
}

Will print out the expected ReferenceError string.

Jerry version:
Checked revision: ac87616f0
Build: debug.linux

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    bugUndesired behaviournormalparserRelated to the JavaScript parser

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