Set last `n` bits in unsigned int

Question

How to set (in most elegant way) exactly n least significant bits of uint32_t? That is to write a function void setbits(uint32_t *x, int n);<

Answer 1

If you meant the least-significant n bits:

((uint32_t)1 << n) - 1

On most architectures, this won't work if n is 32, so you may have to make a special case for that:

n == 32 ? 0xffffffff : (1 << n) - 1

On a 64-bit architecture, a (probably) faster solution is to cast up then down:

(uint32_t)(((uint64_t)1 << n) - 1)

In fact, this might even be faster on a 32-bit architecture since it avoids branching.

Answer 2

const uint32_t masks[33] = {0x0, 0x1, 0x3, 0x7 ...

void setbits(uint32_t *x, int n)
{
   *x |= masks[n];
}

Answer 3

If you mean the most significant n bits:

-1 ^ ((1 << (32 - n)) - 1)

Answer 4

((((1 << (n - 1)) - 1) << 1) | 1)

Set last n bits. n must be > 0. Work with n = 32.

Answer 5

The other answers don't handle the special case of n == 32 (shifting by greater than or equal to the type's width is UB), so here's a better answer:

(uint32_t)(((uint64_t)1 << n) - 1)

Alternatively:

(n == 32) ? 0xFFFFFFFF : (((uint32_t)1 << n) - 1)

Answer 6

Here's a method that doesn't require any arithmetic:

~(~0u << n)