How does multi-byte char '\377777' work?

Question

Clang mentioned this thingy to me when I was trying some various non-std ways to find std::numeric_limits<size_t>::max() for my allocator::max_size()

The expression size seems limited at 4 bytes so it looks pretty much like an impl-defined multi-byte char but I don't know how it works exactly.

printf("0x%016hhX\n", '\377');     // 0x00000000000000FF
printf("0x%016X  \n", '\3777');    // 0x000000000000FF37
printf("0x%016X  \n", '\37777');   // 0x0000000000FF3737
printf("0x%016X  \n", '\377777');  // 0x00000000FF373737 [-Wfour-char-constants]
printf("0x%016X  \n", '\3777777'); // 0x0000000037373737 [ too long ]
printf("0x%016X  \n", '\3777777'); // 0x0000000037373737 [ too long ]

Line 2 - 4 all seem return int but only 4th line warn of -Wfour-char-constants if turn on -Weveryting. Why does 4-byte char have to warn its user?

Please explain how it works. Does it have any colloquial or probably normative term?

godbolt.org/g/7AR9nw

Answer 1

The behaviour on using more than 3 digits with an octal constant specified by \ and enclosed in single quotation characters is not defined by the C++ standard. (Interestingly, you can define multicharacter constants using more than 2 hexadecimal digits, or even notation like 'ab' but the value of the int you get is implementation defined.)

Consult your compiler documentation for the treatment of a longer octal constant.