# 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.

I use this to split string by a delimiter. The first puts the results in a pre-constructed vector, the second returns a new vector.

```
#include <string>
#include <sstream>
#include <vector>
#include <iterator>
template <typename Out>
void split(const std::string &s, char delim, Out result) {
std::istringstream iss(s);
std::string item;
while (std::getline(iss, item, delim)) {
*result++ = item;
}
}
std::vector<std::string> split(const std::string &s, char delim) {
std::vector<std::string> elems;
split(s, delim, std::back_inserter(elems));
return elems;
}
```

Note that this solution does not skip empty tokens, so the following will find 4 items, one of which is empty:

```
std::vector<std::string> x = split("one:two::three", ':');
```

## Best Solution

I can suggest 2 choices to solve the problem, although neither performs a real update.

Use the

`priority_queue`

and push element each time you would like to update it. Accept the fact that you will have useless entries in the queue. When popping the top value, check if it contains the up-to-date value. If not, ignore it and pop the next.This way you delay the removal of the updated element until it comes to the top. I noticed this approach being used by top programmers realizing Dijkstra algorithm.

Use

`set`

. It is also sorted so you are able to extract the greatest element in logarithmic time. You are also able to remove the outdated element before inserting it again. So still no update operation possible, but removal and reinsertion is doable.Seems like the complexity of both approaches is the same.