Thread: sence in garbage out

  1. #1
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    sence in garbage out

    Again i have the following
    Code:
    long unsigned int StringtoNum( char [] );
    
    int main()
    {
        //declare variables
        const char cNumbers[4][51] = {
        "37107287533902102798797998220837590246510135740250",
        "46376937677490009712648124896970078050417018260538",
        "74324986199524741059474233309513058123726617309629",
        "91942213363574161572522430563301811072406154908250",
         };
    
        int i, j, k, iNumCarry;
        long unsigned int iTotal = 0, iTenDigits[4] = { 0 };
        char cStrtoNum[11] = { '\0' };
    
        for ( i = 0; i < 4; i++ )
        {
            for ( j = 40; j < 51; j++ )
            {
                cStrtoNum[j-40] = cNumbers[i][j];
            }
            iTenDigits[i] = StringtoNum( cStrtoNum );
            printf("%s %lu\n", cStrtoNum, iTenDigits[i]);
        }
    
        return 0;
    }
    
    long unsigned int StringtoNum( char strConvert[] )
    {
        long unsigned int iNumConverted = 0;
    
        iNumConverted = atoi( strConvert );
    
        return iNumConverted;
    }
    the output is as follows
    Code:
    0135740250 135740250
    7018260538 18446744072137877562
    6617309629 18446744071736926653
    6154908250 1859940954
    
    Process returned 0 (0x0)   execution time : 0.001 s
    Press ENTER to continue.
    as you can see the first number works fine the rest are utter garbage

  2. #2
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    Code:
    char cStrtoNum[11] = { '\0' };
    I would try

    Code:
    char cStrtoNum[13] = { '\0' };
    Because I think your for loop is going one too much.

    Note: That would mean
    Code:
    char cStrtoNum[12] = { '\0' };
    Would be needed I did one extra in case I am wrong and your are off by two.

    Tim S.
    "...a computer is a stupid machine with the ability to do incredibly smart things, while computer programmers are smart people with the ability to do incredibly stupid things. They are,in short, a perfect match.." Bill Bryson

  3. #3
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    atoi outputs an int, not an unsigned long.
    Your values are too big for a 32-bit int and since atoi is stupid it just wraps around, potentially to a negative value representation which is then interpreted as an unsigned long, giving a large value. (The negative value is essentially added to 2**64, assuming 64-bit unsigned longs.)
    In the last case it wrapped around to a positive value.
    Note that 6154908250 % (2**31) = 1859940954 (2**31 is the largest positive 32-bit int.)
    The behavior is of course undefined, so it doesn't have to work this way. However, it typically works this way.

    atol outputs a long (but not an unsigned long).
    strtoul outputs an unsigned long, so that's the one to use here.
    Code:
    #include <stdio.h>
    #include <stdlib.h>
     
    int main() {
        const char *numbers[4] = {
            "37107287533902102798797998220837590246510135740250",
            "46376937677490009712648124896970078050417018260538",
            "74324986199524741059474233309513058123726617309629",
            "91942213363574161572522430563301811072406154908250",
        };
        unsigned long tenDigits[4] = {0};
     
        for (int i = 0; i < 4; ++i) {
            tenDigits[i] = strtoul(numbers[i] + 40, NULL, 10);
            printf("%s %10lu\n", numbers[i] + 40, tenDigits[i]);
        }
     
        return 0;
    }
    There's more detail for properly handling errors using strtoul:
    strtoul, strtoull - cppreference.com
    Last edited by john.c; 05-26-2023 at 05:51 PM. Reason: Changed ^ to ** for exponentiation.
    A little inaccuracy saves tons of explanation. - H.H. Munro

  4. #4
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    wow thanks for that it help tremendously. The task was to add 100 50 digit numbers and display the first 10 digits of the answer. So i could test and for easy posting on here i just used the first 4 hence the #define N. this is the final solution.
    Code:
    #include <stdio.h>
    #include <stdlib.h>
    
    #define N 4
    
    long unsigned int StringtoNum( char [] );
    void DivideString( const char * [], char [], int, int, int );
    
    int main()
    {
        //declare variables
        const char *cNumbers[N] = {
        "37107287533902102798797998220837590246510135740250",
        "46376937677490009712648124896970078050417018260538",
        "74324986199524741059474233309513058123726617309629",
        "91942213363574161572522430563301811072406154908250",
        };
    
        int i, iNumCarry = 0, j = 50;
        long unsigned int iDivide =10000000000, iFinalTotal = 0;
        double tmpTotal = 0;
        char cStrtoNum[11] = { '\0' };
    
        do
        {
            j -= 10;
            tmpTotal += iNumCarry;
    
            if ( j > 0 )
            {
                for ( i = 0; i < N; i++ )
                {
                    DivideString( cNumbers, cStrtoNum, i, j, j + 11 );
                    tmpTotal += StringtoNum( cStrtoNum );
                }
    
                iNumCarry = tmpTotal / iDivide;
                printf("%lf %d\n", tmpTotal, iNumCarry);
                tmpTotal = 0;
            }
            else
            {
                for ( i = 0; i < N; i++ )
                {
                    DivideString( cNumbers, cStrtoNum, i, j, j + 11 );
                    tmpTotal += StringtoNum( cStrtoNum );
                }
            }
    
        } while ( j > 0 );
    
        iFinalTotal = tmpTotal / 100;
        printf("%lu\n", iFinalTotal);
    
        return 0;
    }
    
    long unsigned int StringtoNum( char strConvert[] )
    {
        long unsigned int iNumConverted = 0;
    
        iNumConverted = strtoul( strConvert, NULL, 10 );
    
        return iNumConverted;
    }
    
    void DivideString( const char *cData[], char strToConvert[], int i, int iLower, int iUpper)
    {
        //declare variable
        int j;
    
        for ( j = iLower; j < iUpper; j++ )
        {
            strToConvert[j - iLower] = cData[i][j];
        }
    
    }
    once again as always thanks for the help

  5. #5
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    Your code can be simplified to the following. Also, at least in the N=4 case, you can't be certain that the final value needs to be divided by 100. If all the numbers started with 1 then it would only need to be divided by 10.

    Note that in modern practice we don't declare all our variables at the top of a function. Instead we declare them where they are first used, in the smallest scope.

    Say "unsigned long", not "long unsigned int".

    It is generally considered idiotic to prefix variables with a type signature.

    Output variables are usually put last in a type signature.
    Code:
    #include <stdio.h>
    #include <stdlib.h>
     
    #define N 4
     
    void substring(const char *in, int from, int to, char *out) {
        int i = from;
        for ( ; i < to; ++i) out[i - from] = in[i];
        out[i - from] = '\0';
    }
     
    int main() {
        const char *numbers[N] = {
            "37107287533902102798797998220837590246510135740250",
            "46376937677490009712648124896970078050417018260538",
            "74324986199524741059474233309513058123726617309629",
            "91942213363574161572522430563301811072406154908250",
        };
        long unsigned total = 0;
     
        for (int j = 40; j >= 0; j -= 10) {
            for (int i = 0; i < N; ++i) {
                char num[11];
                substring(numbers[i], j, j + 11, num);
                total += strtoul(num, NULL, 10);
            }
            if (j > 0)
                total /= 10000000000UL;
        }
     
        while (total >= 10000000000UL) total /= 10;
        printf("%lu\n", total);
     
        return 0;
    }
    You could replace the substring function with an sprintf call:
    Code:
    sprintf(num, "%.10s", numbers[i] + j);
    A little inaccuracy saves tons of explanation. - H.H. Munro

  6. #6
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    i tried your suggestion of having variables with the smallest scope on a different problem and for 5 for loops i had to declare i 4 times. is this the correct practice or if i know i'm going to use a variable multiple times for multiple for loops or the like should i just declare it once and be done with it

  7. #7
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    is this the correct practice
    Yes. Remember to do it like this:
    Code:
    for (int i = 0; i < 10; ++i) ...
    A little inaccuracy saves tons of explanation. - H.H. Munro

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