Strictly speaking, according to the C standard, are unsigned suffixes ever required for numeric literals in any Bitwise operation (to avoid UB)? Particularly around shifts, I think it is undefined behaviour when involving signed numbers and my understanding is that literals are signed but I am not sure. I.e.
4 << x; // or is 4U << x required?
4 >> x;
4 & x;
4 | x;
// etc. for all bit ops
x << 4;
x >> 4;
// etc.
where 4 is a placeholder for any literal.
I know that most compilers would not care either way, I am just interested in what the spec says.
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No, unsigned suffixed aren’t strictly required using bitwise operators, that’s because the ANSI C standard says at chapter 2.9 that bitwise operators must be applied to signed or unsigned integer operands (char, short, int and long).
However, especially in shifts, where the amount of bits to shift cannot be a negative number, would a bit more correct to specify it anyway.
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3
The C standard has no prohibition on using an integer literal (which the standard calls a “constant”) without an unsigned-suffix. However, getting the desired result may require a suffix.
Most simply, consider 1 << n where n is 31 and int is 32 bits. Then 1 << n is not defined by the C standard, because 231 is not representable in int, but 1u << n is defined.
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