C left shift on 64 bits fail

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Because 1 is an int, 32 bits, so (1 << 27)*27 overflows. Use 1ull.

Regarding your comment, if x is a uint64_t, then 1 << x is still an int, but for the multiplication it would be cast to uint64_t, so there’d be no overflow. However, if x >= 31, 1 << x would be undefined behaviour (as the resulting value cannot be represented by a signed 32 bit integer type).

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