Hi Chris,
On 20 August 2017 at 14:05, Christopher Li <sparse@xxxxxxxxxxx> wrote:
> On Sun, Aug 20, 2017 at 8:31 AM, Dibyendu Majumdar
> <mobile@xxxxxxxxxxxxxxx> wrote:
>> I am looking at the parse tree for following snippet:
>>
>> void main(int argc, const char *argv[])
>> {
>> 5;
>> 6.5;
>> "hello";
>> }
>>
>> I see that 5 is treated as EXPR_VALUE, 6.5 as EXPR_FVALUE, but "hello"
>> is treated as EXPR_SYMBOL. Why is that?
>
> Because "hello" is array of char from type point of view.
>
> It is same as
> char no_ident [] = {'h', 'e', 'l', 'l', 'o', '\0'};
>
> except it does not have the name(sym->ident) for this symbol.
>
I see - so every string literal results in an un-named symbol, whose
initializer is set to the string expression?
Regards
Dibyendu
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