在调用函数时要小心

string gen_random(const int len) {
static const char alphanum[] = "0123456789"
        "ABCDEFGHIJKLMNOPQRSTUVWXYZ";

stringstream ss;

for (int i = 0; i < len; ++i) {
    ss << alphanum[rand() % (sizeof(alphanum) - 1)];
}
return ss.str();
}

(改编自@Ates Goral)每次都会产生相同的字符序列。使用

srand(time(NULL));

在调用函数之前，rand()函数总是以1 @kjfletch作为种子。

例如:

void SerialNumberGenerator() {

    srand(time(NULL));
    for (int i = 0; i < 5; i++) {
        cout << gen_random(10) << endl;
    }
}

我的2p解:

#include <random>
#include <string>

std::string random_string(std::string::size_type length)
{
    static auto& chrs = "0123456789"
        "abcdefghijklmnopqrstuvwxyz"
        "ABCDEFGHIJKLMNOPQRSTUVWXYZ";

    thread_local static std::mt19937 rg{std::random_device{}()};
    thread_local static std::uniform_int_distribution<std::string::size_type> pick(0, sizeof(chrs) - 2);

    std::string s;

    s.reserve(length);

    while(length--)
        s += chrs[pick(rg)];

    return s;
}

#include <iostream>
#include <string>
#include <stdlib.h>
int main()
{
    int size;
    std::cout << "Enter size : ";
    std::cin >> size;
    std::string str;
    for (int i = 0; i < size; i++)
    {
        auto d = rand() % 26 + 'a';
        str.push_back(d);
    }
    for (int i = 0; i < size; i++)
    {
        std::cout << str[i] << '\t';
    }

    return 0;
}

Mehrdad Afshari的回答可以解决这个问题，但我觉得对于这个简单的任务来说，它有点太啰嗦了。查询表有时可以创造奇迹:

#include <ctime>
#include <iostream>
#include <unistd.h>

std::string gen_random(const int len) {
    static const char alphanum[] =
        "0123456789"
        "ABCDEFGHIJKLMNOPQRSTUVWXYZ"
        "abcdefghijklmnopqrstuvwxyz";
    std::string tmp_s;
    tmp_s.reserve(len);

    for (int i = 0; i < len; ++i) {
        tmp_s += alphanum[rand() % (sizeof(alphanum) - 1)];
    }
    
    return tmp_s;
}

int main(int argc, char *argv[]) {
    srand((unsigned)time(NULL) * getpid());     
    std::cout << gen_random(12) << "\n";        
    return 0;
}

注意rand生成的随机数质量很差。

下面是我用c++ 11改编的Ates Goral的答案。我在这里添加了lambda，但原理是你可以传入它，从而控制你的字符串包含什么字符:

std::string random_string( size_t length )
{
    auto randchar = []() -> char
    {
        const char charset[] =
        "0123456789"
        "ABCDEFGHIJKLMNOPQRSTUVWXYZ"
        "abcdefghijklmnopqrstuvwxyz";
        const size_t max_index = (sizeof(charset) - 1);
        return charset[ rand() % max_index ];
    };
    std::string str(length,0);
    std::generate_n( str.begin(), length, randchar );
    return str;
}

下面是一个传递lambda到随机字符串函数的例子:http://ideone.com/Ya8EKf

为什么要用c++ 11呢?

因为您可以为感兴趣的字符集生成遵循特定概率分布(或分布组合)的字符串。 因为它内置了对非确定性随机数的支持 因为它支持unicode，所以你可以把它改成国际化的版本。

例如:

#include <iostream>
#include <vector>
#include <random>
#include <functional> //for std::function
#include <algorithm>  //for std::generate_n

typedef std::vector<char> char_array;

char_array charset()
{
    //Change this to suit
    return char_array( 
    {'0','1','2','3','4',
    '5','6','7','8','9',
    'A','B','C','D','E','F',
    'G','H','I','J','K',
    'L','M','N','O','P',
    'Q','R','S','T','U',
    'V','W','X','Y','Z',
    'a','b','c','d','e','f',
    'g','h','i','j','k',
    'l','m','n','o','p',
    'q','r','s','t','u',
    'v','w','x','y','z'
    });
};    

// given a function that generates a random character,
// return a string of the requested length
std::string random_string( size_t length, std::function<char(void)> rand_char )
{
    std::string str(length,0);
    std::generate_n( str.begin(), length, rand_char );
    return str;
}

int main()
{
    //0) create the character set.
    //   yes, you can use an array here, 
    //   but a function is cleaner and more flexible
    const auto ch_set = charset();

    //1) create a non-deterministic random number generator      
    std::default_random_engine rng(std::random_device{}());

    //2) create a random number "shaper" that will give
    //   us uniformly distributed indices into the character set
    std::uniform_int_distribution<> dist(0, ch_set.size()-1);

    //3) create a function that ties them together, to get:
    //   a non-deterministic uniform distribution from the 
    //   character set of your choice.
    auto randchar = [ ch_set,&dist,&rng ](){return ch_set[ dist(rng) ];};

    //4) set the length of the string you want and profit!        
    auto length = 5;
    std::cout<<random_string(length,randchar)<<std::endl;
    return 0;
}

样例输出。

一些更简单和更基本的东西，如果你想让你的字符串包含任何可打印的字符:

#include <time.h>   // we'll use time for the seed
#include <string.h> // this is for strcpy

void randomString(int size, char* output) // pass the destination size and the destination itself
{
    srand(time(NULL)); // seed with time

    char src[size];
    size = rand() % size; // this randomises the size (optional)

    src[size] = '\0'; // start with the end of the string...

    // ...and work your way backwards
    while(--size > -1)
        src[size] = (rand() % 94) + 32; // generate a string ranging from the space character to ~ (tilde)

    strcpy(output, src); // store the random string
}