Grabbing 'n' binary bits from end of unsigned int in C? Bit masking?

-1

I'm pretty new to C and I seem to be messing up my bitmasking. From what I understand, it's a way to grab or create something of a subset from a binary value.

Say I want to grab the last 8 bits of an unsigned int value, containing 00001111000011110000111100001111.

How would I use AND/OR to grab those last 8?

asked Nov 19 '18 at 17:37

RadiantBytes

1

what have you tried so far ? where are your C code attemtps ?

– LoneWanderer
Nov 19 '18 at 17:43

1

Possible duplicate of Get last n bits of binary

– Govind Parmar
Nov 19 '18 at 17:52

add a comment |

-1

I'm pretty new to C and I seem to be messing up my bitmasking. From what I understand, it's a way to grab or create something of a subset from a binary value.

Say I want to grab the last 8 bits of an unsigned int value, containing 00001111000011110000111100001111.

How would I use AND/OR to grab those last 8?

asked Nov 19 '18 at 17:37

RadiantBytes

1

what have you tried so far ? where are your C code attemtps ?

– LoneWanderer
Nov 19 '18 at 17:43

1

Possible duplicate of Get last n bits of binary

– Govind Parmar
Nov 19 '18 at 17:52

add a comment |

-1

I'm pretty new to C and I seem to be messing up my bitmasking. From what I understand, it's a way to grab or create something of a subset from a binary value.

Say I want to grab the last 8 bits of an unsigned int value, containing 00001111000011110000111100001111.

How would I use AND/OR to grab those last 8?

asked Nov 19 '18 at 17:37

RadiantBytes

I'm pretty new to C and I seem to be messing up my bitmasking. From what I understand, it's a way to grab or create something of a subset from a binary value.

Say I want to grab the last 8 bits of an unsigned int value, containing 00001111000011110000111100001111.

How would I use AND/OR to grab those last 8?

c binary hex bit bitmask

asked Nov 19 '18 at 17:37

RadiantBytes

asked Nov 19 '18 at 17:37

RadiantBytes

asked Nov 19 '18 at 17:37

RadiantBytes

asked Nov 19 '18 at 17:37

RadiantBytes

asked Nov 19 '18 at 17:37

RadiantBytes

1

what have you tried so far ? where are your C code attemtps ?

– LoneWanderer
Nov 19 '18 at 17:43

1

Possible duplicate of Get last n bits of binary

– Govind Parmar
Nov 19 '18 at 17:52

add a comment |

1

what have you tried so far ? where are your C code attemtps ?

– LoneWanderer
Nov 19 '18 at 17:43

1

Possible duplicate of Get last n bits of binary

– Govind Parmar
Nov 19 '18 at 17:52

what have you tried so far ? where are your C code attemtps ?

– LoneWanderer
Nov 19 '18 at 17:43

Possible duplicate of Get last n bits of binary

– Govind Parmar
Nov 19 '18 at 17:52

add a comment |

4 Answers
4

active

oldest

votes

Here's a more general solution to generate the mask based on how many bits you're interested int.

unsigned int last_n_bits(unsigned int value, int n)

{

    unsigned int mask = -1;

    if (n < sizeof(unsigned) * CHAR_BIT)

        mask = ((1<<n)-1);

    return value & mask;

}

edited Nov 19 '18 at 18:09

answered Nov 19 '18 at 17:50

cleblanc

3,4611913

you should assert that n < sizeof(unsigned) * CHAR_BIT

– Swordfish
Nov 19 '18 at 17:55

and then you've got a problem getting all bits.

– Swordfish
Nov 19 '18 at 18:03

@Swordfish good valid points.

– cleblanc
Nov 19 '18 at 18:11

The width of unsigned might technically be less than sizeof(unsigned)*CHAR_BIT as there could be padding bits, but dealing with that efficiently would lengthen your answer for little practical benefit.

– PSkocik
Nov 19 '18 at 18:28

add a comment |

You can use the BINARY AND operator for do a binary mask:

unsigned char last_eight_bits = my_number & 0b11111111

answered Nov 19 '18 at 17:43

iElden

684517

2

No need for the bit operation if the target is a char (or another type that is (usually) 8 bits wide). Also 8 times 1 is harder to count than two times f.

– Swordfish
Nov 19 '18 at 17:45

add a comment |

Mask-out all bits > 0xff:

value & 0xffu

answered Nov 19 '18 at 17:44

Swordfish

9,52711436

Or create a mask using (1 << 8) - 1 where 8 represents the number of bits to grab (beware not to exceed the size of the integer, of course).

– Maarten Bodewes
Nov 19 '18 at 17:50

add a comment |

Say I want to grab the last 8 bits of an unsigned int value, containing
00001111000011110000111100001111.

How would I use AND/OR to grab those last 8?

In this case, you would use AND. The following code will grab the 8 least significant bits:

unsigned int number = 0x0F0F0F0Fu;

unsigned int mask = 0x000000FFu; // set all bits to "1" which you want to grab

unsigned int result = number & mask; // result will be 0x0000000F

The &-Operator is used for an AND-Operation with each bit:

    00001111000011110000111100001111

AND 00000000000000000000000011111111

------------------------------------

    00000000000000000000000000001111

Be aware that "0 AND X = 0" and that "1 AND X = X". You can do further investigation at: https://en.wikipedia.org/wiki/Boolean_algebra.

answered Nov 19 '18 at 18:18

GoldenArrows777

311

add a comment |

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4 Answers
4

active

oldest

votes

4 Answers
4

active

oldest

votes

Here's a more general solution to generate the mask based on how many bits you're interested int.

unsigned int last_n_bits(unsigned int value, int n)

{

    unsigned int mask = -1;

    if (n < sizeof(unsigned) * CHAR_BIT)

        mask = ((1<<n)-1);

    return value & mask;

}

edited Nov 19 '18 at 18:09

answered Nov 19 '18 at 17:50

cleblanc

3,4611913

you should assert that n < sizeof(unsigned) * CHAR_BIT

– Swordfish
Nov 19 '18 at 17:55

and then you've got a problem getting all bits.

– Swordfish
Nov 19 '18 at 18:03

@Swordfish good valid points.

– cleblanc
Nov 19 '18 at 18:11

The width of unsigned might technically be less than sizeof(unsigned)*CHAR_BIT as there could be padding bits, but dealing with that efficiently would lengthen your answer for little practical benefit.

– PSkocik
Nov 19 '18 at 18:28

add a comment |

Here's a more general solution to generate the mask based on how many bits you're interested int.

unsigned int last_n_bits(unsigned int value, int n)

{

    unsigned int mask = -1;

    if (n < sizeof(unsigned) * CHAR_BIT)

        mask = ((1<<n)-1);

    return value & mask;

}

edited Nov 19 '18 at 18:09

answered Nov 19 '18 at 17:50

cleblanc

3,4611913

you should assert that n < sizeof(unsigned) * CHAR_BIT

– Swordfish
Nov 19 '18 at 17:55

and then you've got a problem getting all bits.

– Swordfish
Nov 19 '18 at 18:03

@Swordfish good valid points.

– cleblanc
Nov 19 '18 at 18:11

The width of unsigned might technically be less than sizeof(unsigned)*CHAR_BIT as there could be padding bits, but dealing with that efficiently would lengthen your answer for little practical benefit.

– PSkocik
Nov 19 '18 at 18:28

add a comment |

Here's a more general solution to generate the mask based on how many bits you're interested int.

unsigned int last_n_bits(unsigned int value, int n)

{

    unsigned int mask = -1;

    if (n < sizeof(unsigned) * CHAR_BIT)

        mask = ((1<<n)-1);

    return value & mask;

}

edited Nov 19 '18 at 18:09

answered Nov 19 '18 at 17:50

cleblanc

3,4611913

Here's a more general solution to generate the mask based on how many bits you're interested int.

unsigned int last_n_bits(unsigned int value, int n)

{

    unsigned int mask = -1;

    if (n < sizeof(unsigned) * CHAR_BIT)

        mask = ((1<<n)-1);

    return value & mask;

}

edited Nov 19 '18 at 18:09

answered Nov 19 '18 at 17:50

cleblanc

3,4611913

edited Nov 19 '18 at 18:09

answered Nov 19 '18 at 17:50

cleblanc

3,4611913

answered Nov 19 '18 at 17:50

cleblanc

3,4611913

answered Nov 19 '18 at 17:50

cleblanc

3,4611913

you should assert that n < sizeof(unsigned) * CHAR_BIT

– Swordfish
Nov 19 '18 at 17:55

and then you've got a problem getting all bits.

– Swordfish
Nov 19 '18 at 18:03

@Swordfish good valid points.

– cleblanc
Nov 19 '18 at 18:11

The width of unsigned might technically be less than sizeof(unsigned)*CHAR_BIT as there could be padding bits, but dealing with that efficiently would lengthen your answer for little practical benefit.

– PSkocik
Nov 19 '18 at 18:28

add a comment |

you should assert that n < sizeof(unsigned) * CHAR_BIT

– Swordfish
Nov 19 '18 at 17:55

and then you've got a problem getting all bits.

– Swordfish
Nov 19 '18 at 18:03

@Swordfish good valid points.

– cleblanc
Nov 19 '18 at 18:11

The width of unsigned might technically be less than sizeof(unsigned)*CHAR_BIT as there could be padding bits, but dealing with that efficiently would lengthen your answer for little practical benefit.

– PSkocik
Nov 19 '18 at 18:28

you should assert that n < sizeof(unsigned) * CHAR_BIT

– Swordfish
Nov 19 '18 at 17:55

and then you've got a problem getting all bits.

– Swordfish
Nov 19 '18 at 18:03

@Swordfish good valid points.

– cleblanc
Nov 19 '18 at 18:11

The width of unsigned might technically be less than sizeof(unsigned)*CHAR_BIT as there could be padding bits, but dealing with that efficiently would lengthen your answer for little practical benefit.

– PSkocik
Nov 19 '18 at 18:28

add a comment |

You can use the BINARY AND operator for do a binary mask:

unsigned char last_eight_bits = my_number & 0b11111111

answered Nov 19 '18 at 17:43

iElden

684517

2

No need for the bit operation if the target is a char (or another type that is (usually) 8 bits wide). Also 8 times 1 is harder to count than two times f.

– Swordfish
Nov 19 '18 at 17:45

add a comment |

You can use the BINARY AND operator for do a binary mask:

unsigned char last_eight_bits = my_number & 0b11111111

answered Nov 19 '18 at 17:43

iElden

684517

2

No need for the bit operation if the target is a char (or another type that is (usually) 8 bits wide). Also 8 times 1 is harder to count than two times f.

– Swordfish
Nov 19 '18 at 17:45

add a comment |

You can use the BINARY AND operator for do a binary mask:

unsigned char last_eight_bits = my_number & 0b11111111

answered Nov 19 '18 at 17:43

iElden

684517

You can use the BINARY AND operator for do a binary mask:

unsigned char last_eight_bits = my_number & 0b11111111

answered Nov 19 '18 at 17:43

iElden

684517

answered Nov 19 '18 at 17:43

iElden

684517

answered Nov 19 '18 at 17:43

iElden

684517

answered Nov 19 '18 at 17:43

iElden

684517

2

No need for the bit operation if the target is a char (or another type that is (usually) 8 bits wide). Also 8 times 1 is harder to count than two times f.

– Swordfish
Nov 19 '18 at 17:45

add a comment |

2

No need for the bit operation if the target is a char (or another type that is (usually) 8 bits wide). Also 8 times 1 is harder to count than two times f.

– Swordfish
Nov 19 '18 at 17:45

No need for the bit operation if the target is a char (or another type that is (usually) 8 bits wide). Also 8 times 1 is harder to count than two times f.

– Swordfish
Nov 19 '18 at 17:45

add a comment |

Mask-out all bits > 0xff:

value & 0xffu

answered Nov 19 '18 at 17:44

Swordfish

9,52711436

Or create a mask using (1 << 8) - 1 where 8 represents the number of bits to grab (beware not to exceed the size of the integer, of course).

– Maarten Bodewes
Nov 19 '18 at 17:50

add a comment |

Mask-out all bits > 0xff:

value & 0xffu

answered Nov 19 '18 at 17:44

Swordfish

9,52711436

Or create a mask using (1 << 8) - 1 where 8 represents the number of bits to grab (beware not to exceed the size of the integer, of course).

– Maarten Bodewes
Nov 19 '18 at 17:50

add a comment |

Mask-out all bits > 0xff:

value & 0xffu

answered Nov 19 '18 at 17:44

Swordfish

9,52711436

Mask-out all bits > 0xff:

value & 0xffu

answered Nov 19 '18 at 17:44

Swordfish

9,52711436

answered Nov 19 '18 at 17:44

Swordfish

9,52711436

answered Nov 19 '18 at 17:44

Swordfish

9,52711436

answered Nov 19 '18 at 17:44

Swordfish

9,52711436

Or create a mask using (1 << 8) - 1 where 8 represents the number of bits to grab (beware not to exceed the size of the integer, of course).

– Maarten Bodewes
Nov 19 '18 at 17:50

add a comment |

Or create a mask using (1 << 8) - 1 where 8 represents the number of bits to grab (beware not to exceed the size of the integer, of course).

– Maarten Bodewes
Nov 19 '18 at 17:50

Or create a mask using (1 << 8) - 1 where 8 represents the number of bits to grab (beware not to exceed the size of the integer, of course).

– Maarten Bodewes
Nov 19 '18 at 17:50

add a comment |

Say I want to grab the last 8 bits of an unsigned int value, containing
00001111000011110000111100001111.

How would I use AND/OR to grab those last 8?

In this case, you would use AND. The following code will grab the 8 least significant bits:

unsigned int number = 0x0F0F0F0Fu;

unsigned int mask = 0x000000FFu; // set all bits to "1" which you want to grab

unsigned int result = number & mask; // result will be 0x0000000F

The &-Operator is used for an AND-Operation with each bit:

    00001111000011110000111100001111

AND 00000000000000000000000011111111

------------------------------------

    00000000000000000000000000001111

Be aware that "0 AND X = 0" and that "1 AND X = X". You can do further investigation at: https://en.wikipedia.org/wiki/Boolean_algebra.

answered Nov 19 '18 at 18:18

GoldenArrows777

311

add a comment |

Say I want to grab the last 8 bits of an unsigned int value, containing
00001111000011110000111100001111.

How would I use AND/OR to grab those last 8?

In this case, you would use AND. The following code will grab the 8 least significant bits:

unsigned int number = 0x0F0F0F0Fu;

unsigned int mask = 0x000000FFu; // set all bits to "1" which you want to grab

unsigned int result = number & mask; // result will be 0x0000000F

The &-Operator is used for an AND-Operation with each bit:

    00001111000011110000111100001111

AND 00000000000000000000000011111111

------------------------------------

    00000000000000000000000000001111

Be aware that "0 AND X = 0" and that "1 AND X = X". You can do further investigation at: https://en.wikipedia.org/wiki/Boolean_algebra.

answered Nov 19 '18 at 18:18

GoldenArrows777

311

add a comment |

Say I want to grab the last 8 bits of an unsigned int value, containing
00001111000011110000111100001111.

How would I use AND/OR to grab those last 8?

In this case, you would use AND. The following code will grab the 8 least significant bits:

unsigned int number = 0x0F0F0F0Fu;

unsigned int mask = 0x000000FFu; // set all bits to "1" which you want to grab

unsigned int result = number & mask; // result will be 0x0000000F

The &-Operator is used for an AND-Operation with each bit:

    00001111000011110000111100001111

AND 00000000000000000000000011111111

------------------------------------

    00000000000000000000000000001111

Be aware that "0 AND X = 0" and that "1 AND X = X". You can do further investigation at: https://en.wikipedia.org/wiki/Boolean_algebra.

answered Nov 19 '18 at 18:18

GoldenArrows777

311

Say I want to grab the last 8 bits of an unsigned int value, containing
00001111000011110000111100001111.

How would I use AND/OR to grab those last 8?

In this case, you would use AND. The following code will grab the 8 least significant bits:

unsigned int number = 0x0F0F0F0Fu;

unsigned int mask = 0x000000FFu; // set all bits to "1" which you want to grab

unsigned int result = number & mask; // result will be 0x0000000F

The &-Operator is used for an AND-Operation with each bit:

    00001111000011110000111100001111

AND 00000000000000000000000011111111

------------------------------------

    00000000000000000000000000001111

Be aware that "0 AND X = 0" and that "1 AND X = X". You can do further investigation at: https://en.wikipedia.org/wiki/Boolean_algebra.

answered Nov 19 '18 at 18:18

GoldenArrows777

311

answered Nov 19 '18 at 18:18

GoldenArrows777

311

answered Nov 19 '18 at 18:18

GoldenArrows777

311

answered Nov 19 '18 at 18:18

GoldenArrows777

311

add a comment |

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