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Write an iterative C++ program to prompt the user to enter a
nonnegative integer n and its base b_1, the program then will
prompt the user to enter the new base b_2 to convert and
display the integer n to b_2 expansion For example: The user
enter 11111010111100 b_1 = 2 b_2 = 8 the program will
convert and display 37274
Solution
#include <iostream>
#include <cstring>
#include <cmath>
#include <stdlib.h>
using std::cout;
using std::cin;
using std::endl;
int to_dec(char value[], int starting_base);
char* from_dec(int value, int ending_base);
int main() {
char value[30];
int starting_base;
int ending_base;
cout << "This program converts from one base to another, so
long as the bases are" << endl
<< "between 2 and 16." << endl
<< endl;
input_numbers:
cout << "Enter the number, then starting base, then ending
base:" << endl;
cin >> value >> starting_base >> ending_base;
if (starting_base < 2 || starting_base > 16 || ending_base < 2 ||
ending_base > 16) {
cout << "Invalid base(s). ";
goto input_numbers;
}
for (int i=0; value[i]; i++) value[i] = toupper(value[i]);
cout << "Base " << ending_base << ": " <<
from_dec(to_dec(value, starting_base), ending_base) << endl;
return 0;
}
int to_dec(char value[], int starting_base) {
char hex[16] = {'0', '1', '2', '3', '4', '5', '6', '7', '8',
'9', 'A', 'B', 'C', 'D', 'E', 'F'};
long int return_value = 0;
unsigned short int digit = 0;
for (short int pos = strlen(value)-1; pos > -1; pos--) {
for (int i=0; i<starting_base; i++) {
if (hex[i] == value[pos]) {
return_value+=i*pow((float)starting_base, digit++);
break;
}
}
}
return return_value;
}
char* from_dec(int value, int ending_base) {
char hex[16] = {'0', '1', '2', '3', '4', '5', '6', '7', '8',
'9', 'A', 'B', 'C', 'D', 'E', 'F'};
char *return_value = (char *)malloc(30);
unsigned short int digit =
(int)ceil(log10((double)(value+1))/log10((double)ending_base))
;
return_value[digit] = 0;
for (; value != 0; value/=ending_base) return_value[--digit] =
hex[value%ending_base];
return return_value;
}

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Write an iterative C++ program to prompt the user to enter a nonnega.docx

  • 1. Write an iterative C++ program to prompt the user to enter a nonnegative integer n and its base b_1, the program then will prompt the user to enter the new base b_2 to convert and display the integer n to b_2 expansion For example: The user enter 11111010111100 b_1 = 2 b_2 = 8 the program will convert and display 37274 Solution #include <iostream> #include <cstring> #include <cmath> #include <stdlib.h> using std::cout; using std::cin; using std::endl; int to_dec(char value[], int starting_base); char* from_dec(int value, int ending_base); int main() { char value[30]; int starting_base; int ending_base; cout << "This program converts from one base to another, so long as the bases are" << endl
  • 2. << "between 2 and 16." << endl << endl; input_numbers: cout << "Enter the number, then starting base, then ending base:" << endl; cin >> value >> starting_base >> ending_base; if (starting_base < 2 || starting_base > 16 || ending_base < 2 || ending_base > 16) { cout << "Invalid base(s). "; goto input_numbers; } for (int i=0; value[i]; i++) value[i] = toupper(value[i]); cout << "Base " << ending_base << ": " << from_dec(to_dec(value, starting_base), ending_base) << endl; return 0; } int to_dec(char value[], int starting_base) { char hex[16] = {'0', '1', '2', '3', '4', '5', '6', '7', '8', '9', 'A', 'B', 'C', 'D', 'E', 'F'}; long int return_value = 0; unsigned short int digit = 0; for (short int pos = strlen(value)-1; pos > -1; pos--) { for (int i=0; i<starting_base; i++) { if (hex[i] == value[pos]) {
  • 3. return_value+=i*pow((float)starting_base, digit++); break; } } } return return_value; } char* from_dec(int value, int ending_base) { char hex[16] = {'0', '1', '2', '3', '4', '5', '6', '7', '8', '9', 'A', 'B', 'C', 'D', 'E', 'F'}; char *return_value = (char *)malloc(30); unsigned short int digit = (int)ceil(log10((double)(value+1))/log10((double)ending_base)) ; return_value[digit] = 0; for (; value != 0; value/=ending_base) return_value[--digit] = hex[value%ending_base]; return return_value; }