**Executive summary:**

```
int a[17];
size_t n = sizeof(a)/sizeof(a[0]);
```

**Full answer:**

To determine the size of your array in bytes, you can use the `sizeof`

operator:

```
int a[17];
size_t n = sizeof(a);
```

On my computer, ints are 4 bytes long, so n is 68.

To determine the number of elements in the array, we can divide
the total size of the array by the size of the array element.
You could do this with the type, like this:

```
int a[17];
size_t n = sizeof(a) / sizeof(int);
```

and get the proper answer (68 / 4 = 17), but if the type of
`a`

changed you would have a nasty bug if you forgot to change
the `sizeof(int)`

as well.

So the preferred divisor is `sizeof(a[0])`

or the equivalent `sizeof(*a)`

, the size of the first element of the array.

```
int a[17];
size_t n = sizeof(a) / sizeof(a[0]);
```

Another advantage is that you can now easily parameterize
the array name in a macro and get:

```
#define NELEMS(x) (sizeof(x) / sizeof((x)[0]))
int a[17];
size_t n = NELEMS(a);
```

# Setting a bit

Use the bitwise OR operator (`|`

) to set a bit.

```
number |= 1UL << n;
```

That will set the `n`

th bit of `number`

. `n`

should be zero, if you want to set the `1`

st bit and so on upto `n-1`

, if you want to set the `n`

th bit.

Use `1ULL`

if `number`

is wider than `unsigned long`

; promotion of `1UL << n`

doesn't happen until after evaluating `1UL << n`

where it's undefined behaviour to shift by more than the width of a `long`

. The same applies to all the rest of the examples.

# Clearing a bit

Use the bitwise AND operator (`&`

) to clear a bit.

```
number &= ~(1UL << n);
```

That will clear the `n`

th bit of `number`

. You must invert the bit string with the bitwise NOT operator (`~`

), then AND it.

# Toggling a bit

The XOR operator (`^`

) can be used to toggle a bit.

```
number ^= 1UL << n;
```

That will toggle the `n`

th bit of `number`

.

# Checking a bit

You didn't ask for this, but I might as well add it.

To check a bit, shift the number n to the right, then bitwise AND it:

```
bit = (number >> n) & 1U;
```

That will put the value of the `n`

th bit of `number`

into the variable `bit`

.

# Changing the *n*th bit to *x*

Setting the `n`

th bit to either `1`

or `0`

can be achieved with the following on a 2's complement C++ implementation:

```
number ^= (-x ^ number) & (1UL << n);
```

Bit `n`

will be set if `x`

is `1`

, and cleared if `x`

is `0`

. If `x`

has some other value, you get garbage. `x = !!x`

will booleanize it to 0 or 1.

To make this independent of 2's complement negation behaviour (where `-1`

has all bits set, unlike on a 1's complement or sign/magnitude C++ implementation), use unsigned negation.

```
number ^= (-(unsigned long)x ^ number) & (1UL << n);
```

or

```
unsigned long newbit = !!x; // Also booleanize to force 0 or 1
number ^= (-newbit ^ number) & (1UL << n);
```

It's generally a good idea to use unsigned types for portable bit manipulation.

or

```
number = (number & ~(1UL << n)) | (x << n);
```

`(number & ~(1UL << n))`

will clear the `n`

th bit and `(x << n)`

will set the `n`

th bit to `x`

.

It's also generally a good idea to not to copy/paste code in general and so many people use preprocessor macros (like the community wiki answer further down) or some sort of encapsulation.

## Best Solution

For the actual size of the pointer:

If you want the length of the string: