generate a random string in C++11?

Question

I need help with generating a random string using C++11.

I don't know how to continue with that, if you can help me please.

#include <random>
char * random_string()
{

        static const char alphabet[] = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz";
    static const MAX_LEN = 32;  //MAX LENGTH OF THE NEW CHAR RETURNED
    int stringLength = sizeof(alphabet)/sizeof(alphabet[0]);

    for (int i = 0; i<=MAX_LEN;i++)
    {
        //now i don't know what i need to do help!
    }

    static const char test[MAX_LEN];

    return test;

}

Are you looking for [`std::shuffle`](http://en.cppreference.com/w/cpp/algorithm/random_shuffle), or do you want to select some characters multiple times? — Bo Persson, Dec 26 '17 at 11:20
Before the for loop: `string word = "";`, within the loop: `word += alphabet[rand()%62];` (62 is the number of letters in your alphabet array). Btw, char[] is pretty much C-style. Go with std::string in C++. — Aziuth, Dec 26 '17 at 11:27
Look at the example for [uniform_int_distribution](http://en.cppreference.com/w/cpp/numeric/random/uniform_int_distribution) and then think about how you could use those random numbers as indexes into your alphabet to pick characters for your string. — Jesper Juhl, Dec 26 '17 at 11:28
@akroma: *don't comment* your own question, but **edit your question** to improve it. — Basile Starynkevitch, Dec 26 '17 at 11:59

Peter · Accepted Answer · 2017-12-26T11:56:41.460

9

Return a std::string rather than a raw char *. Populate the string as needed to start, and then shuffle it.

For example;

#include <random>
#include <string>

std::string random_string()
{
     std::string str("0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz");

     std::random_device rd;
     std::mt19937 generator(rd());

     std::shuffle(str.begin(), str.end(), generator);

     return str.substr(0, 32);    // assumes 32 < number of characters in str         
}

If you really need to extract a raw const char * from a std::string use its c_str() member function.

int main()
{
    std::string rstr = random_string();

    some_func_that_needs_const_char_pointer(rstr.c_str());
}

edited Dec 26 '17 at 11:56

answered Dec 26 '17 at 11:33

Peter

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the return str.substring(0,32); gives me a error: – akroma Dec 26 '17 at 11:38
10

`shuffle` may not be the correct option here: result string will have no duplicate characters, which possibly is not the requirement. – Gian Paolo Dec 26 '17 at 11:41
The string can have dupplicates, btw the substring gives me a error, why? – akroma Dec 26 '17 at 11:43
Typo. `substring()` should be `substr()`. Fixed. Yes, this approach does not duplicate characters in the returned string. If you want to allow duplicated characters, look up `std::generate()`. – Peter Dec 26 '17 at 11:56

score 7 · Answer 2 · edited Feb 23 '19 at 05:13

#include <random>
using namespace std;

string generate(int max_length){
    string possible_characters = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz";
    random_device rd;
    mt19937 engine(rd());
    uniform_int_distribution<> dist(0, possible_characters.size()-1);
    string ret = "";
    for(int i = 0; i < max_length; i++){
        int random_index = dist(engine); //get index between 0 and possible_characters.size()-1
        ret += possible_characters[random_index];
    }
    return ret;
}

Haseeb Mir · Answer 3 · 2018-05-27T21:05:25.263

7

Using Mersene Twister 19937 generator and Uniform discrete distribution you can generate random strings from ranges like "A-Z","a-z" or "0-9" easily.

#include <iostream>
#include <string>
#include <random>

using namespace std;
int main() {
    mt19937 generator{random_device{}()};
   //modify range according to your need "A-Z","a-z" or "0-9" or whatever you need.
    uniform_int_distribution<int> distribution{'a', 'z'};

    auto generate_len = 6; //modify length according to your need
    string rand_str(generate_len, '\0');
    for(auto& dis: rand_str)
        dis = distribution(generator);

    cout << "Random string generated : " << rand_str << endl;
}

edited May 27 '18 at 21:05

answered May 27 '18 at 20:56

Haseeb Mir

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How can I include my own characters without range, for example I only want to generate a random string with my custom characters `abc012-_`, do i have to do it like this? `distribution{'a', 'b', 'c', '0', '1', '2', '-', '_'}` – Felierix Jan 27 '22 at 09:51
@Felierix no you cannot do like that you have to give range like **A-Z,a-z,0-9** – Haseeb Mir Jan 28 '22 at 11:49
Oh so you can do like `distribution{'a', 'z', '0', '9'}` right? – Felierix Feb 01 '22 at 14:26
@Felierix no uniform distribution only takes 2 Arguments so it can be either **A-Z** or **0-9** – Haseeb Mir Feb 01 '22 at 14:53
@Felierix: To generate a random string with your own custom characters, you need to modify the range of the uniform_int_distribution object to include only the characters you want to use. – Haseeb Mir Mar 03 '23 at 21:50

generate a random string in C++11?

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