How to generate an 8-bit random number in C>

-1

int main(){



    uint8_t *wdata = NULL;

    wdata = calloc(5, sizeof(uint8_t));



    for (int j =0;j<5;j++){



        wdata[j] = rand();



    }



}

The rand() function generates 16 bits of data. How do I generate 8 bits of random values? Do I need to use a custom function for the same?

edited Nov 19 '18 at 16:07

Jonathan Leffler

561k896661018

asked Nov 19 '18 at 16:04

ShivaniSarin

1

So rand returns 16bits, just use 8 of it and discard the rest.
– Kamil Cuk
Nov 19 '18 at 16:10

3

What you've got will assign the least significant 8 bits of the 16-bit value that you say is returned by rand(). Many versions of rand() will return 32-bit values, but that's a separate discussion. The only problem with using the least significant 8 bits is that they may not be as random as you'd like — they might fail various tests of randomness. How much that matters depends on what you're going to do next. Most likely, you can simply use the code shown. Failing that, you probably need a better pseudo-random number generator (PRNG) than rand().
– Jonathan Leffler
Nov 19 '18 at 16:10

1

@JonathanLeffler If you discard some of the random value by a power of 2, it'll still be an even distribution of randomness, won't it? e.g. rand() % 16, rand() % 32, rand() % 64, etc.
– Fiddling Bits
Nov 19 '18 at 16:19

1

See stackoverflow.com/questions/11418113/… and especially links from it, such as to the paper by Marsaglia (sp?).
– Jonathan Leffler
Nov 19 '18 at 16:48

3

@fiddling: if rand is implemented with a.linear congruential generator, then the period of rand%2i is 2i, which is not very random for small i.
– rici
Nov 19 '18 at 18:32

|
show 1 more comment

-1

int main(){



    uint8_t *wdata = NULL;

    wdata = calloc(5, sizeof(uint8_t));



    for (int j =0;j<5;j++){



        wdata[j] = rand();



    }



}

The rand() function generates 16 bits of data. How do I generate 8 bits of random values? Do I need to use a custom function for the same?

edited Nov 19 '18 at 16:07

Jonathan Leffler

561k896661018

asked Nov 19 '18 at 16:04

ShivaniSarin

1

So rand returns 16bits, just use 8 of it and discard the rest.
– Kamil Cuk
Nov 19 '18 at 16:10

3

What you've got will assign the least significant 8 bits of the 16-bit value that you say is returned by rand(). Many versions of rand() will return 32-bit values, but that's a separate discussion. The only problem with using the least significant 8 bits is that they may not be as random as you'd like — they might fail various tests of randomness. How much that matters depends on what you're going to do next. Most likely, you can simply use the code shown. Failing that, you probably need a better pseudo-random number generator (PRNG) than rand().
– Jonathan Leffler
Nov 19 '18 at 16:10

1

@JonathanLeffler If you discard some of the random value by a power of 2, it'll still be an even distribution of randomness, won't it? e.g. rand() % 16, rand() % 32, rand() % 64, etc.
– Fiddling Bits
Nov 19 '18 at 16:19

1

See stackoverflow.com/questions/11418113/… and especially links from it, such as to the paper by Marsaglia (sp?).
– Jonathan Leffler
Nov 19 '18 at 16:48

3

@fiddling: if rand is implemented with a.linear congruential generator, then the period of rand%2i is 2i, which is not very random for small i.
– rici
Nov 19 '18 at 18:32

|
show 1 more comment

-1

int main(){



    uint8_t *wdata = NULL;

    wdata = calloc(5, sizeof(uint8_t));



    for (int j =0;j<5;j++){



        wdata[j] = rand();



    }



}

The rand() function generates 16 bits of data. How do I generate 8 bits of random values? Do I need to use a custom function for the same?

edited Nov 19 '18 at 16:07

Jonathan Leffler

561k896661018

asked Nov 19 '18 at 16:04

ShivaniSarin

int main(){



    uint8_t *wdata = NULL;

    wdata = calloc(5, sizeof(uint8_t));



    for (int j =0;j<5;j++){



        wdata[j] = rand();



    }



}

The rand() function generates 16 bits of data. How do I generate 8 bits of random values? Do I need to use a custom function for the same?

c random types

edited Nov 19 '18 at 16:07

Jonathan Leffler

561k896661018

asked Nov 19 '18 at 16:04

ShivaniSarin

edited Nov 19 '18 at 16:07

Jonathan Leffler

561k896661018

asked Nov 19 '18 at 16:04

ShivaniSarin

edited Nov 19 '18 at 16:07

Jonathan Leffler

561k896661018

edited Nov 19 '18 at 16:07

Jonathan Leffler

561k896661018

edited Nov 19 '18 at 16:07

Jonathan Leffler

561k896661018

asked Nov 19 '18 at 16:04

ShivaniSarin

asked Nov 19 '18 at 16:04

ShivaniSarin

asked Nov 19 '18 at 16:04

ShivaniSarin

1

So rand returns 16bits, just use 8 of it and discard the rest.
– Kamil Cuk
Nov 19 '18 at 16:10

3

What you've got will assign the least significant 8 bits of the 16-bit value that you say is returned by rand(). Many versions of rand() will return 32-bit values, but that's a separate discussion. The only problem with using the least significant 8 bits is that they may not be as random as you'd like — they might fail various tests of randomness. How much that matters depends on what you're going to do next. Most likely, you can simply use the code shown. Failing that, you probably need a better pseudo-random number generator (PRNG) than rand().
– Jonathan Leffler
Nov 19 '18 at 16:10

1

@JonathanLeffler If you discard some of the random value by a power of 2, it'll still be an even distribution of randomness, won't it? e.g. rand() % 16, rand() % 32, rand() % 64, etc.
– Fiddling Bits
Nov 19 '18 at 16:19

1

See stackoverflow.com/questions/11418113/… and especially links from it, such as to the paper by Marsaglia (sp?).
– Jonathan Leffler
Nov 19 '18 at 16:48

3

@fiddling: if rand is implemented with a.linear congruential generator, then the period of rand%2i is 2i, which is not very random for small i.
– rici
Nov 19 '18 at 18:32

|
show 1 more comment

1

So rand returns 16bits, just use 8 of it and discard the rest.
– Kamil Cuk
Nov 19 '18 at 16:10

3

What you've got will assign the least significant 8 bits of the 16-bit value that you say is returned by rand(). Many versions of rand() will return 32-bit values, but that's a separate discussion. The only problem with using the least significant 8 bits is that they may not be as random as you'd like — they might fail various tests of randomness. How much that matters depends on what you're going to do next. Most likely, you can simply use the code shown. Failing that, you probably need a better pseudo-random number generator (PRNG) than rand().
– Jonathan Leffler
Nov 19 '18 at 16:10

1

@JonathanLeffler If you discard some of the random value by a power of 2, it'll still be an even distribution of randomness, won't it? e.g. rand() % 16, rand() % 32, rand() % 64, etc.
– Fiddling Bits
Nov 19 '18 at 16:19

1

See stackoverflow.com/questions/11418113/… and especially links from it, such as to the paper by Marsaglia (sp?).
– Jonathan Leffler
Nov 19 '18 at 16:48

3

@fiddling: if rand is implemented with a.linear congruential generator, then the period of rand%2i is 2i, which is not very random for small i.
– rici
Nov 19 '18 at 18:32

So rand returns 16bits, just use 8 of it and discard the rest.
– Kamil Cuk
Nov 19 '18 at 16:10

What you've got will assign the least significant 8 bits of the 16-bit value that you say is returned by rand(). Many versions of rand() will return 32-bit values, but that's a separate discussion. The only problem with using the least significant 8 bits is that they may not be as random as you'd like — they might fail various tests of randomness. How much that matters depends on what you're going to do next. Most likely, you can simply use the code shown. Failing that, you probably need a better pseudo-random number generator (PRNG) than rand().
– Jonathan Leffler
Nov 19 '18 at 16:10

@JonathanLeffler If you discard some of the random value by a power of 2, it'll still be an even distribution of randomness, won't it? e.g. rand() % 16, rand() % 32, rand() % 64, etc.
– Fiddling Bits
Nov 19 '18 at 16:19

See stackoverflow.com/questions/11418113/… and especially links from it, such as to the paper by Marsaglia (sp?).
– Jonathan Leffler
Nov 19 '18 at 16:48

@fiddling: if rand is implemented with a.linear congruential generator, then the period of rand%2i is 2i, which is not very random for small i.
– rici
Nov 19 '18 at 18:32

|
show 1 more comment

2 Answers
2

active

oldest

votes

How to generate an 8 bit random number in C

Given code such as

uint8_t *wdata = calloc( 5, sizeof( uint8_t ) );

something like

for ( int j = 0;j < 5; j++ )

{

    wdata[ j ] = rand();

}

will work fine.

Integer assignment to an unsigned value truncates, which is exactly what you want. See 6.5.16.1 Simple assignment of the C Standard.

Note that RAND_MAX is guaranteed by 7.22.2 Pseudo-random sequence generation functions, paragraph 5 to be at least 32767, which is "wider" than 8 bits, so the result is guaranteed to "fill up" an 8-bit variable.

edited Nov 19 '18 at 17:44

Jonathan Leffler

561k896661018

answered Nov 19 '18 at 16:10

Andrew Henle

19.5k21333

Yes, but, c-faq.com/lib/notveryrand.html. See c-faq.com/lib/randrange.html.
– Neil Edelman
Nov 19 '18 at 19:58

1

@NeilEdelman A poor PRNG is a poor PRNG, and when the modulo value is much less than the maximum result from the PRNG the modulo operation itself won't add any measurable bias to the result.
– Andrew Henle
Nov 20 '18 at 17:38

add a comment |

Because 8 bit number can contain maximum 0-255. so if i divide rand()%256 we get 8 bit random number.

  int main(){

    uint8_t *wdata = NULL;

    wdata = calloc(5, sizeof(uint8_t));

    for (int j =0;j<5;j++){

        wdata[j] = rand()%256;

   }

}

for more randomness you use some prime number add or subtract difference after dividing from 256.

answered Nov 19 '18 at 16:42

Susheel Dwivedi

554

Since they are unsigned 8 bit numbers, rand()%256 and rand() produce the same thing; although, it may be some benefit for readability.
– Neil Edelman
Nov 19 '18 at 19:14

rand(void) returns a pseudo-random number in the range of 0 to RAND_MAX. RAND_MAX is a constant whose default value may vary between implementations but it is granted to be at least 32767.
– Susheel Dwivedi
Nov 19 '18 at 19:24

add a comment |

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

How to generate an 8 bit random number in C

Given code such as

uint8_t *wdata = calloc( 5, sizeof( uint8_t ) );

something like

for ( int j = 0;j < 5; j++ )

{

    wdata[ j ] = rand();

}

will work fine.

Integer assignment to an unsigned value truncates, which is exactly what you want. See 6.5.16.1 Simple assignment of the C Standard.

edited Nov 19 '18 at 17:44

Jonathan Leffler

561k896661018

answered Nov 19 '18 at 16:10

Andrew Henle

19.5k21333

Yes, but, c-faq.com/lib/notveryrand.html. See c-faq.com/lib/randrange.html.
– Neil Edelman
Nov 19 '18 at 19:58

1

@NeilEdelman A poor PRNG is a poor PRNG, and when the modulo value is much less than the maximum result from the PRNG the modulo operation itself won't add any measurable bias to the result.
– Andrew Henle
Nov 20 '18 at 17:38

add a comment |

How to generate an 8 bit random number in C

Given code such as

uint8_t *wdata = calloc( 5, sizeof( uint8_t ) );

something like

for ( int j = 0;j < 5; j++ )

{

    wdata[ j ] = rand();

}

will work fine.

Integer assignment to an unsigned value truncates, which is exactly what you want. See 6.5.16.1 Simple assignment of the C Standard.

edited Nov 19 '18 at 17:44

Jonathan Leffler

561k896661018

answered Nov 19 '18 at 16:10

Andrew Henle

19.5k21333

Yes, but, c-faq.com/lib/notveryrand.html. See c-faq.com/lib/randrange.html.
– Neil Edelman
Nov 19 '18 at 19:58

1

@NeilEdelman A poor PRNG is a poor PRNG, and when the modulo value is much less than the maximum result from the PRNG the modulo operation itself won't add any measurable bias to the result.
– Andrew Henle
Nov 20 '18 at 17:38

add a comment |

How to generate an 8 bit random number in C

Given code such as

uint8_t *wdata = calloc( 5, sizeof( uint8_t ) );

something like

for ( int j = 0;j < 5; j++ )

{

    wdata[ j ] = rand();

}

will work fine.

Integer assignment to an unsigned value truncates, which is exactly what you want. See 6.5.16.1 Simple assignment of the C Standard.

edited Nov 19 '18 at 17:44

Jonathan Leffler

561k896661018

answered Nov 19 '18 at 16:10

Andrew Henle

19.5k21333

How to generate an 8 bit random number in C

Given code such as

uint8_t *wdata = calloc( 5, sizeof( uint8_t ) );

something like

for ( int j = 0;j < 5; j++ )

{

    wdata[ j ] = rand();

}

will work fine.

Integer assignment to an unsigned value truncates, which is exactly what you want. See 6.5.16.1 Simple assignment of the C Standard.

edited Nov 19 '18 at 17:44

Jonathan Leffler

561k896661018

answered Nov 19 '18 at 16:10

Andrew Henle

19.5k21333

edited Nov 19 '18 at 17:44

Jonathan Leffler

561k896661018

edited Nov 19 '18 at 17:44

Jonathan Leffler

561k896661018

edited Nov 19 '18 at 17:44

Jonathan Leffler

561k896661018

answered Nov 19 '18 at 16:10

Andrew Henle

19.5k21333

answered Nov 19 '18 at 16:10

Andrew Henle

19.5k21333

answered Nov 19 '18 at 16:10

Andrew Henle

19.5k21333

Yes, but, c-faq.com/lib/notveryrand.html. See c-faq.com/lib/randrange.html.
– Neil Edelman
Nov 19 '18 at 19:58

1

@NeilEdelman A poor PRNG is a poor PRNG, and when the modulo value is much less than the maximum result from the PRNG the modulo operation itself won't add any measurable bias to the result.
– Andrew Henle
Nov 20 '18 at 17:38

add a comment |

Yes, but, c-faq.com/lib/notveryrand.html. See c-faq.com/lib/randrange.html.
– Neil Edelman
Nov 19 '18 at 19:58

1

@NeilEdelman A poor PRNG is a poor PRNG, and when the modulo value is much less than the maximum result from the PRNG the modulo operation itself won't add any measurable bias to the result.
– Andrew Henle
Nov 20 '18 at 17:38

Yes, but, c-faq.com/lib/notveryrand.html. See c-faq.com/lib/randrange.html.
– Neil Edelman
Nov 19 '18 at 19:58

@NeilEdelman A poor PRNG is a poor PRNG, and when the modulo value is much less than the maximum result from the PRNG the modulo operation itself won't add any measurable bias to the result.
– Andrew Henle
Nov 20 '18 at 17:38

add a comment |

Because 8 bit number can contain maximum 0-255. so if i divide rand()%256 we get 8 bit random number.

  int main(){

    uint8_t *wdata = NULL;

    wdata = calloc(5, sizeof(uint8_t));

    for (int j =0;j<5;j++){

        wdata[j] = rand()%256;

   }

}

for more randomness you use some prime number add or subtract difference after dividing from 256.

answered Nov 19 '18 at 16:42

Susheel Dwivedi

554

Since they are unsigned 8 bit numbers, rand()%256 and rand() produce the same thing; although, it may be some benefit for readability.
– Neil Edelman
Nov 19 '18 at 19:14

rand(void) returns a pseudo-random number in the range of 0 to RAND_MAX. RAND_MAX is a constant whose default value may vary between implementations but it is granted to be at least 32767.
– Susheel Dwivedi
Nov 19 '18 at 19:24

add a comment |

Because 8 bit number can contain maximum 0-255. so if i divide rand()%256 we get 8 bit random number.

  int main(){

    uint8_t *wdata = NULL;

    wdata = calloc(5, sizeof(uint8_t));

    for (int j =0;j<5;j++){

        wdata[j] = rand()%256;

   }

}

for more randomness you use some prime number add or subtract difference after dividing from 256.

answered Nov 19 '18 at 16:42

Susheel Dwivedi

554

Since they are unsigned 8 bit numbers, rand()%256 and rand() produce the same thing; although, it may be some benefit for readability.
– Neil Edelman
Nov 19 '18 at 19:14

rand(void) returns a pseudo-random number in the range of 0 to RAND_MAX. RAND_MAX is a constant whose default value may vary between implementations but it is granted to be at least 32767.
– Susheel Dwivedi
Nov 19 '18 at 19:24

add a comment |

Because 8 bit number can contain maximum 0-255. so if i divide rand()%256 we get 8 bit random number.

  int main(){

    uint8_t *wdata = NULL;

    wdata = calloc(5, sizeof(uint8_t));

    for (int j =0;j<5;j++){

        wdata[j] = rand()%256;

   }

}

for more randomness you use some prime number add or subtract difference after dividing from 256.

answered Nov 19 '18 at 16:42

Susheel Dwivedi

554

Because 8 bit number can contain maximum 0-255. so if i divide rand()%256 we get 8 bit random number.

  int main(){

    uint8_t *wdata = NULL;

    wdata = calloc(5, sizeof(uint8_t));

    for (int j =0;j<5;j++){

        wdata[j] = rand()%256;

   }

}

for more randomness you use some prime number add or subtract difference after dividing from 256.

answered Nov 19 '18 at 16:42

Susheel Dwivedi

554

answered Nov 19 '18 at 16:42

Susheel Dwivedi

554

answered Nov 19 '18 at 16:42

Susheel Dwivedi

554

answered Nov 19 '18 at 16:42

Susheel Dwivedi

554

Since they are unsigned 8 bit numbers, rand()%256 and rand() produce the same thing; although, it may be some benefit for readability.
– Neil Edelman
Nov 19 '18 at 19:14

rand(void) returns a pseudo-random number in the range of 0 to RAND_MAX. RAND_MAX is a constant whose default value may vary between implementations but it is granted to be at least 32767.
– Susheel Dwivedi
Nov 19 '18 at 19:24

add a comment |

Since they are unsigned 8 bit numbers, rand()%256 and rand() produce the same thing; although, it may be some benefit for readability.
– Neil Edelman
Nov 19 '18 at 19:14

rand(void) returns a pseudo-random number in the range of 0 to RAND_MAX. RAND_MAX is a constant whose default value may vary between implementations but it is granted to be at least 32767.
– Susheel Dwivedi
Nov 19 '18 at 19:24

Since they are unsigned 8 bit numbers, rand()%256 and rand() produce the same thing; although, it may be some benefit for readability.
– Neil Edelman
Nov 19 '18 at 19:14

rand(void) returns a pseudo-random number in the range of 0 to RAND_MAX. RAND_MAX is a constant whose default value may vary between implementations but it is granted to be at least 32767.
– Susheel Dwivedi
Nov 19 '18 at 19:24

add a comment |

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