Question

Purpose

To develop C++ code that incorporates: (1) generalized conversion from one number base to another; (2) string character extraction and insertion; and (3) a loop trifecta (while, for, and do loops). After this exercise you should better understand the difference between integer and string versions of a number, and the difference between integer and printable character versions of a single digit. Conversions between string and integer representations of a number are needed since you can't perform arithmetic directly on the string version of a number, and you can't index the printable digits directly with the integer version of a number.

What to do

Create a new Visual Studio console project named assignment06. Write C++ code inside the main() function that first prompts for two integers in the range 2 to 10. The first value is the base for numbers entered on the console (ibase), and the second value is the base for numbers displayed on the console (obase). Like the Linux bc program, your program allows the ibase and obase to be independently set. For example, ibase can be 2 and obase can be 10, or vice versa. The ibase and obase can be the same (not very useful but good for testing), or one could be 5 (the valid digits are 0 to 4) and the other could be 8 (the valid digits are 0 to 7).

While the ibase and obase can be read into integer variables, the number to be converted from the ibase to the obase should first be read in as a string, converted to an integer according to the ibase, and then be converted to a output string according to the obase. While the ibase and obase values are checked for validity, the number's input string does not have to be checked for validity.

Algorithm

Here is an appropriate algorithm expressed in pseudocode.

   prompt for ibase and obase
   if (ibase and obase are valid) {
     declare a string named input
     prompt for input 
     while (input != "x") {
       int value = 0
       for (i from 0 to (input.length() - 1)) {
         convert the input[i] printable character to an integer named digit_int
         value = (value * ibase) + digit_int
       }
       declare a string named output
       do {
         int remainder = value % obase
         convert the integer remainder to a printable character named digit_char
         prepend digit_char to output
         value = value / obase
       } while (value != 0)
       display ibase, input, obase, output
       prompt for a string named input
     }
   }
   else {
     display a message indicating that ibase and/or obase is invalid
   }

Converting the algorithm to C++

Pseudocode does not need to precisely specify how an operation is performed in a particular programming language. Examples in the above pseudocode are:

1. The characters in a C++ string are indexed from 0 (leftmost character) to n-1 (rightmost character) where n is the length of the string. Strings are initially empty (n = 0). There is no string character with index n. Hence a string character index must range from 0 to n-1.
2. Converting a printable (ASCII) digit to a binary integer value. Take a look at an ASCII table such as https://www.ascii-code.com/ (Links to an external site.) The binary code for the character '0' is 0x30 which is not the same as the binary value for 0 (0x00). The ASCII code for the character '1' is 0x31, '2' is 0x32, ... , '9' is 0x39. Hence one needs to subtract 0x30 (or just '0') to convert from printable ASCII to an integer. It may seem strange but the C/C++ char type is treated as an 8-bit integer allowing you to subtract '0' from another character. Using '0' is preferred over the constant 0x30 because it is clearer what is going on and you don't need to remember the code.
3. Converting an integer (binary) digit value to a printable (ASCII) character is the inverse function, so just add '0' to the integer value and assign to a char variable.
4. While a character can be appended to a string using the append() member function, the insert() member function can be used to prepend a character. You can read about how to use insert() at https://stackoverflow.com/questions/3223302/c-insert-char-to-a-string (Links to an external site.) As explained, output.insert(0, 1, digit_char) inserts 1 character (digit_char) at the beginning (index 0) of the string.
5. There is no need for float or double variables or arithmetic.

here is what i have done so far but im sure there are some mistakes. plz help me complete it

#include <iostream>
#include <cstring>

int main()
{
int ibase, obase;
std::cout << "enter the value for the base of the input followed by the value of the base for the output";
std::cin >> ibase >> obase;
if (ibase <= 10 && ibase >= 2 && obase <= 10 && obase >= 2) {
std::string input;
std::cout << "enter a interger that you want to convert";
std::cin >> input;
while (input != "x") {
int value = 0;
int n;
n = size(input);
for (int i = 0; i < n; i++) {
int digit_int;
  
digit_int = input[i];

  
  

}

}

}
}

Answer 1

#include <iostream>
#include <cstring>

int main()
{
int ibase, obase;
std::cout << "enter the value for the base of the input followed by the value of the base for the output\n";
std::cin >> ibase >> obase;
if (ibase <= 10 && ibase >= 2 && obase <= 10 && obase >= 2) {
std::string input;
std::cout << "enter a interger that you want to convert\n";
std::cin >> input;
while (input != "x") {
int value = 0;
int n;
n = input.length();
int digit_int;
for (int i = 0; i < n; i++) {
digit_int = (int)input[i] - '0';
value=(value*ibase) + digit_int;
}
std::string output;
do{
int remainde = value % obase;
char digit_char = (char)remainde + '0';
output.insert(0, 1,digit_char);
value = value/obase;
}
while(value!=0);
std::cout <<ibase<<" "<<input<<" "<<obase<<" "<<output;
std::cout << "\nenter a interger that you want to convert\n";
std::cin >> input;
}
}
else
{
std::cout<<"Invalid";
}
}

I did code just as things were mentioned in the pseudocode and things that you have researched. One mistake that i saw was size(input). It will be input.length() to find the length of the string.

Hope all problems are solved by this code. For any doubts or questions comment below.