I need help with the following problem:

Write an expression that results in the least significant byte of an integer x being set to 0, but all other bits unchanged. For example: 0x98234493 becomes 0x98234400.

I am confused on how to access the least significant byte.

```
public static int Q3(int x) {
return (x & 0xFFFFFFFF);
}
```

What am i doing wrong?

asked on Stack Overflow Sep 5, 2017 by Utsab

You could do

```
public static int Q3(int x) {
return (x & ~255);
}
```

255 in binary would have all the bits in the least significant byte set while all others are zero.

Therefore ~255, which is the complement of 255, will have all bits except those in the least significant byte set while all bits of the least significant byte are zeros.

This ~255 can be `&`

ed with the input number `n`

to set all bits in the least significant byte to zero while the other bits remain unchanged.

answered on Stack Overflow Sep 9, 2017 by J...S

The solution will be
```
public static int Q3(int x) {
return (x & OxF);
}
```

answered on Stack Overflow Jan 30, 2018 by Omolola Olamide

If all you want is the least significant byte of an integer (as your question is stated), then it's very simple:

`byte b = (byte) (x & 0xFF);`

This works because all the high bits of `0xFF`

will be set to 0 by default, so you are masking out everything but the bottom byte. A byte is 8 bits and each 'F' is 4 bits (written in hexadecimal). In binary, an F is 1111, so FF masks out everything but the bottom 8 bits.

answered on Stack Overflow Sep 29, 2019 by EntangledLoops

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