Babbage's number with C++ not showing proper results

I was given a question to find babbage's number (A number whose square ends with 269696, for example 99736 squared is 9947269696, thus being a Babbage Number). My implementation in C++ is provided below. It is not providing the desired results. The first result is correct, but the subsequent ones are not. I probably think that it has to do with casting, or the variables getting out of range. But upon checking with the range of unsigned long long int , I can't understand what is wrong with my code.

#include<iostream>



using namespace std;



int main(int argc, char **argv){



    unsigned long long square = 0;



    for(int i = 519; i < 100000; i++){



        square = i*i;    



        if(square % 1000000 == 269696)

            cout<<"Number: "<<i<<", Square: "<<square<<endl;



    }



}

This is the output I'm getting :

Number: 25264, Square: 638269696

Number: 119328, Square: 1354269696

Number: 158816, Square: 18446744073162269696

Number: 186992, Square: 606269696

Number: 188008, Square: 987269696

Number: 331424, Square: 18446744071882269696

Number: 439080, Square: 18446744073227269696

I do see that the squares are also inaccurate, but I don't understand why.

Here's the list of Babbage's number

25264
99736
150264
224736
275264
349736
400264
474736
525264
599736
650264
724736
775264
849736
900264
974736

asked Nov 21 '18 at 6:56

Anshul Kumar

add a comment |

#include<iostream>



using namespace std;



int main(int argc, char **argv){



    unsigned long long square = 0;



    for(int i = 519; i < 100000; i++){



        square = i*i;    



        if(square % 1000000 == 269696)

            cout<<"Number: "<<i<<", Square: "<<square<<endl;



    }



}

This is the output I'm getting :

Number: 25264, Square: 638269696

Number: 119328, Square: 1354269696

Number: 158816, Square: 18446744073162269696

Number: 186992, Square: 606269696

Number: 188008, Square: 987269696

Number: 331424, Square: 18446744071882269696

Number: 439080, Square: 18446744073227269696

I do see that the squares are also inaccurate, but I don't understand why.

Here's the list of Babbage's number

25264
99736
150264
224736
275264
349736
400264
474736
525264
599736
650264
724736
775264
849736
900264
974736

asked Nov 21 '18 at 6:56

Anshul Kumar

add a comment |

#include<iostream>



using namespace std;



int main(int argc, char **argv){



    unsigned long long square = 0;



    for(int i = 519; i < 100000; i++){



        square = i*i;    



        if(square % 1000000 == 269696)

            cout<<"Number: "<<i<<", Square: "<<square<<endl;



    }



}

This is the output I'm getting :

Number: 25264, Square: 638269696

Number: 119328, Square: 1354269696

Number: 158816, Square: 18446744073162269696

Number: 186992, Square: 606269696

Number: 188008, Square: 987269696

Number: 331424, Square: 18446744071882269696

Number: 439080, Square: 18446744073227269696

I do see that the squares are also inaccurate, but I don't understand why.

Here's the list of Babbage's number

25264
99736
150264
224736
275264
349736
400264
474736
525264
599736
650264
724736
775264
849736
900264
974736

asked Nov 21 '18 at 6:56

Anshul Kumar

#include<iostream>



using namespace std;



int main(int argc, char **argv){



    unsigned long long square = 0;



    for(int i = 519; i < 100000; i++){



        square = i*i;    



        if(square % 1000000 == 269696)

            cout<<"Number: "<<i<<", Square: "<<square<<endl;



    }



}

This is the output I'm getting :

Number: 25264, Square: 638269696

Number: 119328, Square: 1354269696

Number: 158816, Square: 18446744073162269696

Number: 186992, Square: 606269696

Number: 188008, Square: 987269696

Number: 331424, Square: 18446744071882269696

Number: 439080, Square: 18446744073227269696

I do see that the squares are also inaccurate, but I don't understand why.

Here's the list of Babbage's number

25264
99736
150264
224736
275264
349736
400264
474736
525264
599736
650264
724736
775264
849736
900264
974736

c++

asked Nov 21 '18 at 6:56

Anshul Kumar

asked Nov 21 '18 at 6:56

Anshul Kumar

asked Nov 21 '18 at 6:56

Anshul Kumar

asked Nov 21 '18 at 6:56

Anshul Kumar

asked Nov 21 '18 at 6:56

Anshul Kumar

add a comment |

3 Answers
3

active

oldest

votes

The variable i is a signed int. Since e.g. 99736 * 99736 is much larger than even an unsigned int can handle on all common platforms you will have arithmetic overflow by the multiplication, and for signed integer types that leads to undefined behavior.

You need to make i an unsigned long long as well.

answered Nov 21 '18 at 7:02

Some programmer dude

303k25265426

add a comment |

You have to change the data type of variable i from signed integer to unsigned integer. You are getting the wrong square of the number because of integer overflow

size_t square = 0;



for(size_t i = 519; i < 100000; i++){

    square = i*i;    

    if(square % 1000000 == 269696)

        cout<<"Number: "<<i<<", Square: "<< square <<"n";

}

As pointed by Some Programmer Dude in the comment, if you are working on 32-bit machine then you need to use unsigned long long otherwise size_t will perfectly on 64-bit machine.

edited Nov 21 '18 at 7:35

answered Nov 21 '18 at 7:04

Shravan40

4,37031632

2

Considering that unsigned int on all common platforms is 32 bits, and the second Babbage number is well above nine billions, that won't be enough.

– Some programmer dude
Nov 21 '18 at 7:09

add a comment |

Make Changes into the type of int i variable unsigned long long i.

As when you are calculating multiplication of int * int for some ranges it's exceed the range of int may results into the undefined behaviour.

#include<iostream>



using namespace std;



int main(int argc, char **argv){



    unsigned long long square = 0;



    for(unsigned long long i = 519; i < 100000; i++){



        square = i*i;    



        if(square % 1000000 == 269696)

            cout<<"Number: "<<i<<", Square: "<<square<<endl;



    }



}

edited Nov 23 '18 at 6:32

answered Nov 21 '18 at 7:05

Mr. Roshan

1,393524

add a comment |

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

You need to make i an unsigned long long as well.

answered Nov 21 '18 at 7:02

Some programmer dude

303k25265426

add a comment |

You need to make i an unsigned long long as well.

answered Nov 21 '18 at 7:02

Some programmer dude

303k25265426

add a comment |

You need to make i an unsigned long long as well.

answered Nov 21 '18 at 7:02

Some programmer dude

303k25265426

You need to make i an unsigned long long as well.

answered Nov 21 '18 at 7:02

Some programmer dude

303k25265426

answered Nov 21 '18 at 7:02

Some programmer dude

303k25265426

answered Nov 21 '18 at 7:02

Some programmer dude

303k25265426

answered Nov 21 '18 at 7:02

Some programmer dude

303k25265426

add a comment |

You have to change the data type of variable i from signed integer to unsigned integer. You are getting the wrong square of the number because of integer overflow

size_t square = 0;



for(size_t i = 519; i < 100000; i++){

    square = i*i;    

    if(square % 1000000 == 269696)

        cout<<"Number: "<<i<<", Square: "<< square <<"n";

}

As pointed by Some Programmer Dude in the comment, if you are working on 32-bit machine then you need to use unsigned long long otherwise size_t will perfectly on 64-bit machine.

edited Nov 21 '18 at 7:35

answered Nov 21 '18 at 7:04

Shravan40

4,37031632

2

Considering that unsigned int on all common platforms is 32 bits, and the second Babbage number is well above nine billions, that won't be enough.

– Some programmer dude
Nov 21 '18 at 7:09

add a comment |

You have to change the data type of variable i from signed integer to unsigned integer. You are getting the wrong square of the number because of integer overflow

size_t square = 0;



for(size_t i = 519; i < 100000; i++){

    square = i*i;    

    if(square % 1000000 == 269696)

        cout<<"Number: "<<i<<", Square: "<< square <<"n";

}

As pointed by Some Programmer Dude in the comment, if you are working on 32-bit machine then you need to use unsigned long long otherwise size_t will perfectly on 64-bit machine.

edited Nov 21 '18 at 7:35

answered Nov 21 '18 at 7:04

Shravan40

4,37031632

2

Considering that unsigned int on all common platforms is 32 bits, and the second Babbage number is well above nine billions, that won't be enough.

– Some programmer dude
Nov 21 '18 at 7:09

add a comment |

You have to change the data type of variable i from signed integer to unsigned integer. You are getting the wrong square of the number because of integer overflow

size_t square = 0;



for(size_t i = 519; i < 100000; i++){

    square = i*i;    

    if(square % 1000000 == 269696)

        cout<<"Number: "<<i<<", Square: "<< square <<"n";

}

As pointed by Some Programmer Dude in the comment, if you are working on 32-bit machine then you need to use unsigned long long otherwise size_t will perfectly on 64-bit machine.

edited Nov 21 '18 at 7:35

answered Nov 21 '18 at 7:04

Shravan40

4,37031632

You have to change the data type of variable i from signed integer to unsigned integer. You are getting the wrong square of the number because of integer overflow

size_t square = 0;



for(size_t i = 519; i < 100000; i++){

    square = i*i;    

    if(square % 1000000 == 269696)

        cout<<"Number: "<<i<<", Square: "<< square <<"n";

}

As pointed by Some Programmer Dude in the comment, if you are working on 32-bit machine then you need to use unsigned long long otherwise size_t will perfectly on 64-bit machine.

edited Nov 21 '18 at 7:35

answered Nov 21 '18 at 7:04

Shravan40

4,37031632

edited Nov 21 '18 at 7:35

answered Nov 21 '18 at 7:04

Shravan40

4,37031632

answered Nov 21 '18 at 7:04

Shravan40

4,37031632

answered Nov 21 '18 at 7:04

Shravan40

4,37031632

2

Considering that unsigned int on all common platforms is 32 bits, and the second Babbage number is well above nine billions, that won't be enough.

– Some programmer dude
Nov 21 '18 at 7:09

add a comment |

2

Considering that unsigned int on all common platforms is 32 bits, and the second Babbage number is well above nine billions, that won't be enough.

– Some programmer dude
Nov 21 '18 at 7:09

Considering that unsigned int on all common platforms is 32 bits, and the second Babbage number is well above nine billions, that won't be enough.

– Some programmer dude
Nov 21 '18 at 7:09

add a comment |

Make Changes into the type of int i variable unsigned long long i.

As when you are calculating multiplication of int * int for some ranges it's exceed the range of int may results into the undefined behaviour.

#include<iostream>



using namespace std;



int main(int argc, char **argv){



    unsigned long long square = 0;



    for(unsigned long long i = 519; i < 100000; i++){



        square = i*i;    



        if(square % 1000000 == 269696)

            cout<<"Number: "<<i<<", Square: "<<square<<endl;



    }



}

edited Nov 23 '18 at 6:32

answered Nov 21 '18 at 7:05

Mr. Roshan

1,393524

add a comment |

Make Changes into the type of int i variable unsigned long long i.

As when you are calculating multiplication of int * int for some ranges it's exceed the range of int may results into the undefined behaviour.

#include<iostream>



using namespace std;



int main(int argc, char **argv){



    unsigned long long square = 0;



    for(unsigned long long i = 519; i < 100000; i++){



        square = i*i;    



        if(square % 1000000 == 269696)

            cout<<"Number: "<<i<<", Square: "<<square<<endl;



    }



}

edited Nov 23 '18 at 6:32

answered Nov 21 '18 at 7:05

Mr. Roshan

1,393524

add a comment |

Make Changes into the type of int i variable unsigned long long i.

As when you are calculating multiplication of int * int for some ranges it's exceed the range of int may results into the undefined behaviour.

#include<iostream>



using namespace std;



int main(int argc, char **argv){



    unsigned long long square = 0;



    for(unsigned long long i = 519; i < 100000; i++){



        square = i*i;    



        if(square % 1000000 == 269696)

            cout<<"Number: "<<i<<", Square: "<<square<<endl;



    }



}

edited Nov 23 '18 at 6:32

answered Nov 21 '18 at 7:05

Mr. Roshan

1,393524

Make Changes into the type of int i variable unsigned long long i.

As when you are calculating multiplication of int * int for some ranges it's exceed the range of int may results into the undefined behaviour.

#include<iostream>



using namespace std;



int main(int argc, char **argv){



    unsigned long long square = 0;



    for(unsigned long long i = 519; i < 100000; i++){



        square = i*i;    



        if(square % 1000000 == 269696)

            cout<<"Number: "<<i<<", Square: "<<square<<endl;



    }



}

#include<iostream>



using namespace std;



int main(int argc, char **argv){



    unsigned long long square = 0;



    for(unsigned long long i = 519; i < 100000; i++){



        square = i*i;    



        if(square % 1000000 == 269696)

            cout<<"Number: "<<i<<", Square: "<<square<<endl;



    }



}

#include<iostream>



using namespace std;



int main(int argc, char **argv){



    unsigned long long square = 0;



    for(unsigned long long i = 519; i < 100000; i++){



        square = i*i;    



        if(square % 1000000 == 269696)

            cout<<"Number: "<<i<<", Square: "<<square<<endl;



    }



}

edited Nov 23 '18 at 6:32

answered Nov 21 '18 at 7:05

Mr. Roshan

1,393524

edited Nov 23 '18 at 6:32

answered Nov 21 '18 at 7:05

Mr. Roshan

1,393524

answered Nov 21 '18 at 7:05

Mr. Roshan

1,393524

answered Nov 21 '18 at 7:05

Mr. Roshan

1,393524

add a comment |

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