making an int a char string

This is a discussion on making an int a char string within the C++ Programming forums, part of the General Programming Boards category; i need help with this program i am currently working on. it invloves 7 random numbers (made into a character ...

  1. #1
    C++No0b!!!
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    making an int a char string

    i need help with this program i am currently working on.
    it invloves 7 random numbers (made into a character string) that could potentially be used for something like a password for instance. example: say the random numbers come up as 1 4 7 6 3 2 8. i want a character string that says: 1476328.
    i tried using strcpy and strcat as well as dabbled in atoi none of which helped me.
    does anyone know how to do this?
    thanks for your help.

  2. #2
    C++No0b!!!
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    i can post an example of code if needbe...

  3. #3
    C++ Witch laserlight's Avatar
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    You could insert the numbers to a stringstream, and then extract from the stringstream to a string:
    Code:
    std::string password;
    const int numbers_size = 7;
    int numbers[numbers_size] = {1, 4, 7, 6, 3, 2, 8};
    std::stringstream ss;
    std::copy(numbers, numbers + numbers_size, std::ostream_iterator<int>(ss));
    ss >> password;
    Note that you'll need the <algorithm>, <iterator>, <string> and <sstream> standard headers included for my example to work.
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  4. #4
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    boost::lexical_cast might just be what you need.

    Definitely more intuitive than strstream if not as efficient

    The code would be something like:
    Code:
    int numbers[7] = {1, 4, 7, 6, 3, 2, 8};
    
    string out = boost::lexical_cast<string>(numbers[0]) + boost::lexical_cast<string>(numbers[1]) + boost::lexical_cast<string>(numbers[2]) + boost::lexical_cast<string>(numbers[3]) + boost::lexical_cast<string>(numbers[4]) + boost::lexical_cast<string>(numbers[5]) + boost::lexical_cast<string>(numbers[6]);
    Of course you coulde also do it in a loop
    Code:
    int numbers[7] = {1, 4, 7, 6, 3, 2, 8};
    string out;
    
    for (int i = 0; i < 7; ++i)
    {
            out += boost::lexical_cast<string>(numbers[i]);
    }
    you can find the boost libraries at www.boost.org

  5. #5
    Registered User manofsteel972's Avatar
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    You could just create a lookup table and use the random number as an index into the string.
    Code:
    string table="0123456789";
    string s;
    
    s+=table[rand()%10];
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  6. #6
    Maz
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    There is a function called atoi() which converst array to integer. It fullfills posix.1 standard, so it HOULD be a part of the standard libraries included in any serious compiler... Unfortunately M$ is not too known of staying within standards... (Well, I think M$ compilers do have that included too though, and gcc definitely has ) Only problem with atoi is, that it does not inform about errors... If there's some lets say alphabets in array, atoi will just return 0.

    EDIT:
    If you need error reporting, use strtol instead

  7. #7
    Maz
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    Oh, sorry. I missread you. atoi and strtol work in opposite way :|

    Well, there is function called itoa(), but it is not included in posix/ansi standards as far as I know... Anyways some compilers do include it in their standard libraries. I also found quite a good one when I had no time to make my own, by using our dear friend google

  8. #8
    Cat without Hat CornedBee's Avatar
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    atoi() is part of the ANSI C standard and thus present in every hosted C implementation, i.e. every compiler a new programmer is every likely to encounter.

    For a straight C way to solve the given problem, there's snprintf (C99, so older compilers like MS's might not have it - MS has something like it, though, you have to look it up.)
    Code:
    #include <string.h>
    
    #define BUFSIZE 100
    
    int main()
    {
      int numbers[] = { 2, 3, 4, 1, 6, 7, 5 };
      char buffer[BUFSIZE];
    
      snprintf(buffer, BUFSIZE, "%i%i%i%i%i%i%i", numbers[0], numbers[1] ...);
    }
    Last edited by CornedBee; 03-29-2006 at 05:53 AM.
    All the buzzt!
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  9. #9
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    >>If there's some lets say alphabets in array, atoi will just return 0.

    Well, not really. atoi() first skips over all white space (spaces, tabs) then converts up to the first non-numeric digit, then returns whatever value it has converted up to that point. Although it is documented to return 0 on error I have never encountered that problem. The main problem with atoi() is that it does not detect data overflow.

  10. #10
    C++ Witch laserlight's Avatar
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    You could just create a lookup table and use the random number as an index into the string.
    The problem is that the OP didnt specify that the random numbers will always be in the range [0, 9]

    EDIT:
    if we did make that assumption, I suppose my example could be modified to:
    Code:
    char int2digit(int n) {
    	return static_cast<char>(n) + '0';
    }
    
    // ...
    
    std::string password;
    const int numbers_size = 7;
    int numbers[numbers_size] = {1, 4, 7, 6, 3, 2, 8};
    std::transform(numbers, numbers + numbers_size, std::back_inserter(password), int2digit);
    Last edited by laserlight; 03-29-2006 at 08:47 AM.
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  11. #11
    C++No0b!!!
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    well i was told atio could be used but i could not make it work. i tried using the strcpy and strcat wich obviously didnt work but i suppose it was worth a try... ill post an example of what i have

  12. #12
    C++No0b!!!
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    here was an attempt using the strcat which does not work since you cant convert an int to a character string.

    Code:
    #include <cstdlib> 
    #include <ctime> 
    #include <iostream>
    #include <string>
    
    using namespace std; 
    
    int main() 
    { 
    srand((unsigned)time(0));
     
    int random_integer, x=0, n1, n2, n3, n4, n5, n6, n7;
    char pass[7];
     
    do
    { 
    random_integer = (rand()%9)+1; 
    cout << random_integer;
    x++;
    
    if (x==1)
    {
    n1=random_integer;
    } 
    
    if (x==2)
    {
    n2=random_integer;
    }
    
    if (x==3)
    {
    n3=random_integer;
    }
    
    if (x==4)
    {
    n4=random_integer;
    }
    
    if (x==5)
    {
    n5=random_integer;
    }
    
    if (x==6)
    {
    n6=random_integer;
    }
    
    if (x==7)
    {
    n7=random_integer;
    }
    }
    while (x!=7);
    
    cout << endl << n1 << n2 << n3 << n4 << n5 << n6 << n7;
    
    strcat(pass,n1);
    strcat(pass,n2);
    strcat(pass,n3);
    strcat(pass,n4);
    strcat(pass,n5);
    strcat(pass,n6);
    strcat(pass,n7);
    
    getch();
    }

  13. #13
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    Code:
    Just Sorry..
    Last edited by Banzai10; 03-30-2006 at 12:02 PM. Reason: Wrong code again

  14. #14
    C++ Witch laserlight's Avatar
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    here was an attempt using the strcat which does not work since you cant convert an int to a character string.
    There are a number of suggestions in this thread, have you not looked at them?

    Banzai10, your example has a terrific bug with it, but you cant see it since you deceived yourself by printing the original integers.
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  15. #15
    C++No0b!!!
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    my output was very strange. maybe i did it incorrectly but it did run...

    Code:
    #include <cstdlib> 
    #include <ctime> 
    #include <iostream>
    #include <string>
    
    using namespace std; 
    
    int main() 
    { 
    srand((unsigned)time(0));
     
    int random_integer, x=0, n1, n2, n3, n4, n5, n6, n7;
    char pass;
    string out;
     
    do
    { 
    random_integer = (rand()%9)+1; 
    cout << random_integer;
    x++;
    
    if (x==1)
    {
    n1=random_integer;
    } 
    
    if (x==2)
    {
    n2=random_integer;
    }
    
    if (x==3)
    {
    n3=random_integer;
    }
    
    if (x==4)
    {
    n4=random_integer;
    }
    
    if (x==5)
    {
    n5=random_integer;
    }
    
    if (x==6)
    {
    n6=random_integer;
    }
    
    if (x==7)
    {
    n7=random_integer;
    }
    }
    while (x!=7);
    
    out += (char)n1;
    out += (char)n2;
    out += (char)n3;
    out += (char)n4;
    out += (char)n5;
    out += (char)n6;
    out += (char)n7;
    
    cout << out;
    
    cin.get();
    }

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