perror - 1 year ago 65

C Question

I want to get a function that will set the

`n`

`1`

`bitmask (5) = 0b11111 = 31`

bitmask (0) = 0

I, first, had this implementation (

`mask_t`

`typedef`

`uint64_t`

`mask_t bitmask (unsigned short n) {`

return ((((mask_t) 1) << n) - 1;

}

Everything is fine except when the function hit

`bitmask (64)`

`mask_t`

`bitmask (64) = 0`

`1`

So, I have two questions:

- Why do I have this behavior ? Pushing the by 64 shifts on the left should clear the register and remain with
`1`

, then applying the`0`

should fill the register with`-1`

s...`1`

- What is the proper way to achieve this function ?

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Answer Source

Yes this is a well known problem. There are easy ways to implement this function over the range 0..63 and over the range 1..64 (see below, #1), but 0..64 is more difficult (see even further below, #2)

Based on inverting, shifting, and inverting again:

```
mask_t bitmask (unsigned short n) {
// pre: n in [1..64], n=0 acts like n=64
mask_t full = - ((mask_t)1);
return ~(full << (n & 63));
}
```

Well of course you can just take either the "straight" or "inverted" mask generation and then special-case the "missing" `n`

,

```
mask_t bitmask (unsigned short n) {
if (n == 64) return -((mask_t)1);
return (((mask_t) 1) << n) - 1;
}
```

This tends to compile to a branch, though it technically doesn't *have* to.

You can do it without `if`

(not tested)

```
mask_t bitmask (unsigned int n) {
mask_t x = (n ^ 64) >> 6;
return (x << (n & 63)) - 1;
}
```

The idea here is that we're going to either shift 1 left by some amount the same as in your original code, or 0 in the case that `n = 64`

. Shifting 0 left by 0 is just going to be 0 again, subtracting 1 sets all 64 bits.

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