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char initialization with unicode withouth the '\u'

 
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Can anyone explain me why this char initialization is allright?


It doesn't use the '\u' convention. Why is it ok?
 
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It's not Unicode, it's octal definition of charactor resolving to 1*64 + 0*8 + 1 = 65, which is ascii 'A' character.
 
Leandro Melo
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Humm... are you sure?
I thought octal literals started with 0. Also why do you need the quotes anyway?
I thought an octal initialization would look like this.

char c = 067;
 
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From the book by Khalid Mughal & Rolf Rasmussen:

We can use the escape sequence \ddd to specify a character literal by octal value, where each digit d can be any octal digit (0-7). The number of digits must be 3 or fewer, and the octal value cannot exceed \377, that is, only the first 256 chracters can be specified with this notation.

e.g.:
'\141', '\46', '\60'
 
Leandro Melo
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Ok then.
 
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