Base conversion is a mini C++-based project which converts the entered number from any base ranging from base 2 to base 10 to any base ranging from base 2 to base 16 numbers.
The project is based on C++ and uses various aspects of C++ to do the computation for base conversion.
You can enter any number from base two to ten to the program and the program outputs the number converted in your required base from base two to base sixteen.
Check out other interesting C++ projects posted on bestengineeringprojects.com
- Telephone Diary using C++
- Employee Management System Using C++
- 3D Ball, graphics-based project on C++
- Color the text [C++ Based]
C++ code for Base Conversion project:
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 |
#include <iostream.h> #include <conio.h> #include <iomanip.h> #include <math.h> int main() { int cotinue=1; long int org_num=0, org_base=0, valid_org_val, pos_org_num, new_num, base10_num, remainder; int new_base=0; int pow_limit, pow_test; int cur_term=0; cout<<setiosflags(ios::left); do { do { cout<<endl<<"Enter the base that the original number must be converted to\n"<<setw(67)<<"(the value should be between 2 and 16, inclusively)."; cin>>new_base; } while((new_base < 2) || (new_base > 16)); do { cout<<"\n\nInput the integer that must be converted.\n\nIf the new base is 2, then the original number must be\ngreater than or equal to -1,000 and less than or equal to 1,000.\n\nIf the new base is > 2 and < 11, then the initial number\nshould be between -10,000 and 10,000, inclusively.\n\nWhen the new base is larger than 10, the number\n"<<setw(67)<<"must be between -1,000,000 and 1,000,000, inclusively."; cin>>org_num; } while((((org_num < -1000) || (org_num > 1000)) && (new_base == 2)) || (((org_num < -10000) || (org_num > 10000)) && ((new_base > 2) && (new_base < 11))) || (((org_num < -1000000) || (org_num > 1000000)) && (new_base > 10))); cout<<endl; do { cout<<"\nWhat is the base of the initial number\n"<<setw(67)<<"(the base should be between 2 and 10, inclusively)?"; cin>>org_base; } while((org_base < 2) || (org_base > 10)); clrscr(); //Second Page Begins to Be Processed //***Full Conversion Process Begins Below if(org_num < 0) pos_org_num = org_num * -1; else pos_org_num = org_num; //**Start of Conversion to Base Ten for(pow_test=0; pow(10, pow_test) <= pos_org_num; ++pow_test); for(pow_limit=(pow_test - 1), remainder=pos_org_num, base10_num=0, valid_org_val=1; pow_limit >= 0; --pow_limit) { cur_term = (remainder / pow(10, pow_limit)); if(cur_term >= org_base) { pow_limit=-1; valid_org_val=0; } base10_num = (base10_num + (cur_term * pow(org_base, pow_limit))); remainder = (remainder - (pow(10, pow_limit) * cur_term)); } //**End of Base Ten Conversion if(valid_org_val == 0) cout<<"\n\n\n\nError: The Base and the Value of the Original Number Do not Coincide"; else { cout<<"\n\n\n\nThe conversion of "<<org_num<<" from base "<<org_base<<" to base "<<new_base<<" reads as follows:\n\n"; for(pow_test=0; pow(new_base, pow_test) <= base10_num; ++pow_test); if(new_base < 11) { for(pow_limit=(pow_test-1), remainder=base10_num, new_num=0; pow_limit >= 0; --pow_limit) { cur_term = pow(new_base, pow_limit); new_num = new_num + (pow(10, pow_limit) * (remainder / cur_term)); remainder = (remainder - ((remainder / cur_term) * cur_term)); } if(org_num < 0) new_num = new_num * -1; cout<<new_num; //The user's final, highly desired conversion solution is outputted onto the screen. } else { for(pow_limit=(pow_test-1), remainder=base10_num; pow_limit >= 0; --pow_limit) { cur_term = pow(new_base, pow_limit); if((remainder / cur_term) < 10) cout<<(remainder / cur_term); else cout<<char(65 + ((remainder / cur_term) - 10)); remainder = (remainder - ((remainder / cur_term) * cur_term)); } } } //***End of Full Base Conversion cout<<"\n\n\n\n\n\n\n\tWould you like to run the conversion process again?\n\t\t\t\t(1 = Yes, 2 = No)\n\t\t\t\t\t"; cin>>cotinue; clrscr(); } while(cotinue == 1); return 0; } |