我想创建一个随机字符串,由字母-数字字符组成。我希望能够指定字符串的长度。

如何在c++中做到这一点?


当前回答

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。

其他回答

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。

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;
}
//C++ Simple Code
#include <bits/stdc++.h>
using namespace std;
int main() {
vector<char> alphanum =
    {'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'
};
string s="";
int len=5;
srand(time(0)); 
for (int i = 0; i <len; i++) {
    int t=alphanum.size()-1;
    int idx=rand()%t;
    s+= alphanum[idx];
}
cout<<s<<" ";
return 0;
}

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;
    }
}

您可以使用random()方法生成一个基本的随机字符串。 下面的代码生成一个由小写字母、大写字母和数字组成的随机字符串。

String randomStrGen(int numChars){
  String genStr="";
  int sizeStr=0;
  
  while(sizeStr<numChars){
    int asciiPos= random(48,122);
    if((asciiPos>57 && asciiPos<65) || (asciiPos>90 && asciiPos<97))
      continue;
    genStr+=(char) asciiPos;
    sizeStr++;
  }
  
  return genStr;
}

如果需要更安全的随机数生成器,只需将random()函数替换为更安全的随机数生成器。

此外,还可以通过将ASCII限制(48,122)更改为另一个自定义值来调整可能生成的字符