Thread: How to generate a random chars in C programming

Hi, How can I generate a function that's printing a random characters of CODE ACII from A-Z uppercase! ?? is there a specific command from printing a random characters?
In every attempt the input must be changed automatically (not stuck on specific character!)


thanks

Do something like this

Code:

srand(time(NULL)); //only do this once
(rand()%(90-65))+65; //65 is ASCII for capital A, 90 is ASCII for capital Z

See this

Here's an example that doesn't even rely on ASCII or even 'A' to 'Z' being consecutive numbers each increasing by one

Code:

#include <stdio.h>
#include <stdlib.h>
#include <time.h>
 
int main(void)
{
    static const char ucase[] = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
    const size_t ucase_count = sizeof(ucase) - 1; // ucase includes terminating '\0'
    size_t i, j;

    srand(time(NULL));

    for(i = 0; i < 10; i++) {
        for(j = 0; j < ucase_count; j++) {
            char random_char;
            int random_index = (double)rand() / RAND_MAX * ucase_count;
            
            random_char = ucase[random_index];
            printf("%c", random_char);
        }
        printf("\n");
    }

    return 0;
}

I leave it as an exercise for the reader to guess why I might have used the strange way of calculating random_index instead of just rand() % ucase_count (Edit: and even if it matters for this use case)

>> int random_index = (double)rand() / RAND_MAX * ucase_count;
I think you're missing a +1 in there.

>> I leave it as an exercise ...
Not using % avoids using just lower order bits from rand(), or something like that.
However, division/multiplication doesn't help with distribution: Preserving RNG distribution on int cast (Nominal Animal has a cool implementation in post #10 as well)

gg

Originally Posted by Codeplug

>> int random_index = (double)rand() / RAND_MAX * ucase_count;
I think you're missing a +1 in there.

The bug is worse than that (I am missing a -1) but the problem is more that rand() does not produce RAND_MAX very often at all and the "target range" is so small. So I would need to do two things: a) add the -1 where it's required; and b) think about the distribution of rand() a bit more

Edit:
In Nominal Animal's post he says "Again, none of this aliasing occurs if you just use min + (int)((double)interval * prng())." but this is not the answer here and after looking at the horrible distribution of C's rand() function (glibc actually) it's not a solution either

Edit 2: Oh I see where you got that I was missing a +1 from, but nah, in the case of the code above I'm missing a -1 because the range is [0, RAND_MAX] and I'm not calculating max - min + 1 as in post #10 of your link (which is why the sign reversal is necessary and I should be multiplying by (ucase_count - 1)

>> In Nominal Animal's post he says "Again, none of this aliasing occurs if you just use min + (int)((double)interval * prng())."
I'm assuming that's a brain fart.

If you look at the end of post #10, the same formula is presented with:
>> Yes, this way will have some aliasing, too ...

This is the "+1" that I was thinking of:

Code:

#include <stdio.h>
#include <stdlib.h>
#include <time.h>

int rand26_A()
{
    return (int)((double)rand() / RAND_MAX * 26.0);
}

int rand26_B()
{
    return (int)((double)rand() / (RAND_MAX + 1) * 26.0);
    //                                        ^
    //                                       here
}

void test(int (*rand26)())
{
    int table[26] = {0};
    int n;
    for (n = 0; n < 100000; ++n)
    {
        int i = rand26();
        if (i < 0 || i > 25)
            printf("Uh Oh\n");
        else
            ++table[i];
    }

    for (n = 0; n < 26; ++n)
        printf("%d\n", table[n]);
    printf("\n\n");
}

int main()
{
    printf("rand26_A:\n");
    srand(26);
    test(&rand26_A);

    printf("rand26_B:\n");
    srand(26);
    test(&rand26_B);

    return 0;
}

Code:

rand26_A:
Uh Oh
Uh Oh
Uh Oh
Uh Oh
Uh Oh
Uh Oh
Uh Oh
3860
3788
3914
3830
3958
3893
3918
3825
3747
3788
3845
3830
3830
3892
3886
3827
3851
3731
3885
3930
3877
3761
3730
3862
3808
3927

rand26_B:
3860
3788
3914
3830
3958
3893
3919
3824
3747
3788
3847
3828
3830
3895
3883
3827
3854
3735
3878
3932
3878
3763
3727
3862
3808
3932

>> Until I figure out a way to get around rand()'s distribution problem
The best you can do to preserve rand's distribution is to use "exclusion method" in post #13 (which is from "Accelerated C++").

gg

Originally Posted by Codeplug

>> In Nominal Animal's post he says "Again, none of this aliasing occurs if you just use min + (int)((double)interval * prng())."

I'm assuming that's a brain fart.

Yes, it probably was. I had a similar fart so it's understandable.

Originally Posted by Codeplug

This is the "+1" that I was thinking of:

Code:

int rand26_B()
{
    return (int)((double)rand() / (RAND_MAX + 1) * 26.0);
    //                                        ^
    //                                       here
}

Yes. I just wasn't thinking when I wrote the code and after you mentioned the bug, dividing by RAND_MAX+1 also entered my mind, but then I decided against it because I *guessed* (shoot me! ) that RAND_MAX might be equivalent to INT_MAX (turns out that on my system, at least, that it is) and I didn't want to deal with potential overflow so I changed my mind and went with subtracting 1 from the required max-value (26) instead.

Getting RAND_MAX to actually be returned by rand() seems to be highly dependent upon the seed used. (based on my admittedly superficial investigation)

Originally Posted by Codeplug

The best you can do to preserve rand's distribution is to use "exclusion method" in post #13 (which is from "Accelerated C++").

Oh I think the best way for me to preserve its distribution is to just not use rand() Quite honestly I haven't used rand() in a long, long time.


Edit: Actually your example does rely on integer overflow. Is that intentional? It certainly helps with the distribution

Yeah. I think the right way to do the non-mod formula is:

Code:

int rand26_B()
{
    return (int)((double)rand() / (RAND_MAX / 26.0 + 1));
}

gg