How to write a random uppercase char generator in C++?

Question

My take on this would be

#include <iostream>

int main()
{
    for( int i = 0; i < 20; ++i) {
        std::cout << (char) ((random() % (int('Z') + 1)) + int('A'))  << '\n';
    }
}

This seems to be very straight forward. You generate a number, that is between everything lower or equal to Z and greater or equal to A. If you play with it, you will find that there are also other chars generated. Why is that?

You probably meant `(char) ((random() % ('Z' - 'A' + 1)) + int('A'))` (which is no portable anyway). — Jarod42, Jan 14 '20 at 13:01
Possible duplicate of https://stackoverflow.com/questions/440133/how-do-i-create-a-random-alpha-numeric-string-in-c — Galik, Jan 14 '20 at 13:03

Jarod42 · Accepted Answer · 2020-01-14T13:55:57.293

8

With <random>, and not assuming ASCII, you might do:

const std::string alphabet = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
constexpr std::size_t output_size = 20;
std::mt19937 rng{std::random_device{}()};
std::uniform_int_distribution<> dis(0, alphabet.size() - 1);
std::string res;
res.reserve(output_size);

std::generate_n(std::back_inserter(res),
                output_size,
                [&](){ return alphabet[dis()]; });

edited Jan 14 '20 at 13:55

answered Jan 14 '20 at 13:03

Jarod42

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There's nothing documented about this, but doesn't the inherit method of how these algorithms work mean your output would always be "ordered" (If A shows up in the sample it will always be first)? – NathanOliver Jan 14 '20 at 13:41
@NathanOliver: Oops, indeed, `std::sample` is not the right tool here. – Jarod42 Jan 14 '20 at 13:56
@NathanOliver Are you referring to the *stable* remark [here](http://eel.is/c++draft/alg.random.sample)? It appears to be so. – Bob__ Jan 14 '20 at 13:56
@Jarod42 Not neccessarly, you just have to shuffle the outputted result. – Bob__ Jan 14 '20 at 13:57
@Bob__ I'm just referring to everything I've ever seen on a single pass sampler like `std::sample`. I've never seen one where the output wasn't ordered the same way the input was. – NathanOliver Jan 14 '20 at 13:58
@Jarod42 I like the change, but mabye make that an addition to your answer or just a new answer instead? All the votes and the accepts are on your old answer, not this new one. – NathanOliver Jan 14 '20 at 13:59
@Bob__: `"AAAAAAAA"` won't be generated with `std::sample`, but it OP attempt's allow that. – Jarod42 Jan 14 '20 at 14:00
@NathanOliver: but my answer with `std::sample` was wrong (even if up-voted)... – Jarod42 Jan 14 '20 at 14:02

Bathsheba · Answer 2 · 2020-01-14T13:07:55.743

5

char c = *("ABCDEFGHIJKLMNOPQRSTUVWXYZ" + rand() % 26);

is at least portable. Note the use of pointer arithmetic on the const char[27] constant which decays to a const char* type in the expression. Switch out rand() for something out of std::random if the above is not sufficiently random.

edited Jan 14 '20 at 13:07

answered Jan 14 '20 at 13:03

Bathsheba

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Possibly `alphabet[rand() % 26];`. – Jarod42 Jan 14 '20 at 13:05
@Jarod42: Yeah, I'm trying to be cute. – Bathsheba Jan 14 '20 at 13:05
Anything which includes `% 26` assumes 26 letters which is only the case in english. – Michael Chourdakis Jan 14 '20 at 13:13
@MichaelChourdakis: Not entirely; I'm assuming a character encoding supported by any C compiler. – Bathsheba Jan 14 '20 at 13:14
1

@MichaelChourdakis: `26` is just the length of provided string. And more languages than just English use alphabet of 26 letters ;-) – Jarod42 Jan 14 '20 at 13:25
1

FWIW, you can make this a little shorter using `char c = "ABCDEFGHIJKLMNOPQRSTUVWXYZ"[rand() % 26];` – NathanOliver Jan 14 '20 at 13:51
1

@NathanOliver that's what Jarod42 suggests in the very first comment. – Aykhan Hagverdili Jan 14 '20 at 13:53

score 3 · Answer 3 · answered Jan 14 '20 at 13:01

Your math is wrong. Assuming ASCII, (random() % (int('Z') + 1) could return anything up to 'Z', including lowercase characters, numbers, unprintable characters and other things. If you add int('A') to that, you got a high change of getting a result that's outside of the uppercase characters, and even over 127 and therefore not ASCII anymore.

You could simply have 'A' + (random() % 26) instead.

score 1 · Answer 4 · answered Jan 14 '20 at 13:37

1

Another option is to use std::shuffle. You can use that to mix up your alphabet and take the first N characters you need from that shuffled string. That would look like

std::string alphabet = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
std::shuffle(alphabet.begin(), alphabet.end(), std::mt19937{std::random_device{}()});
std::cout << alphabet.substr(20); // N in this case

answered Jan 14 '20 at 13:37

NathanOliver

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I like this one too, noting of course that it's impossible in this scheme to draw the same digit twice, you're late to the party. Have an upvote! – Bathsheba Jan 14 '20 at 13:46
@Bathsheba Better late then never ;) Thanks. Yeah, wasn't sure if uniqueness was required or not. – NathanOliver Jan 14 '20 at 13:47

score 0 · Answer 5 · answered Jan 14 '20 at 13:01

'Z' equals to 90, so the code:

(random() % (int('Z') + 1)

Generates numbers between 1 and 90. Then you add 'A' which is 65, so your numbers are in range 66 - 155. To fix it you have to change your code to:

std::cout << (char) ((random() % (int('Z' - 'A') + 1)) + int('A'))  << '\n';

score -1 · Answer 6 · answered Jan 14 '20 at 14:24

-1

Well, I would write it this way. It just samples a RNG in the range of 'A' to 'Z', and then casts it to char.

std::mt19937 rng{ std::random_device{}() };
std::uniform_int_distribution<> dis('A', 'Z');
std::cout << static_cast<char>(dis(rng));

answered Jan 14 '20 at 14:24

kookie

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How to write a random uppercase char generator in C++?

6 Answers6