C++ function convertCtoD - c++

Question

I'm new to C++. As part of an assignment we have to write to functions, but I don't know what the teacher mean by what he is requesting. Has anyone seen this or at least point me in the right direction. I don't want you to write the function, I just don't know what the output or what he is asking. I'm actually clueless right now.
Thank you
convertCtoD( )
This function is sent a null terminated character array
where each character represents a Decimal (base 10) digit.
The function returns an integer which is the base 10 representation of the characters.
convertBtoD( )
This function is sent a null terminated character array
where each character represents a Binary (base 2) digit.
The function returns an integer which is the base 10 representation of the character.

Answer 1

This function is sent a null terminated character array where each character represents a Decimal (base 10) digit. The function returns an integer which is the base 10 representation of the characters.
I'll briefly mention the fact that "an integer which is the base 10 representation of the characters" is useless here, the integer will represent the value whereas "base 10 representation" is the presentation of said value.
However, the desription given simply means you take in a (C-style) string of digits and put out an integer. So you would start with:
int convertCtoD(char *decimalString) {
int retVal = 0
// TBD: needs actual implementation.
return retVal;
}
This function is sent a null terminated character array where each character represents a Binary (base 2) digit. The function returns an integer which is the base 10 representation of the character.
This will be very similar:
int convertBtoD(char *binaryString) {
int retVal = 0
// TBD: needs actual implementation.
return retVal;
}
You'll notice I've left the return type as signed even though there's no need to handle signed values at all. You'll see why in the example implementation I provide below as I'm using it to return an error condition. The reason I'm providing code even though you didn't ask for it is that I think five-odd years is enough of a gap to ensure you can't cheat by passing off my code as your own :-)
Perhaps the simplest example would be:
int convertCToD(char *decimalStr) {
// Initialise accumulator to zero.
int retVal = 0;
// Process each character.
while (*str != '\0') {
// Check character for validity, add to accumulator (after
// converting char to int) then go to next character.
if ((*str < '0') || (*str > '9')) return -1;
retVal *= 10;
retVal += *str++ - '0';
}
return retVal;
}
The binary version would basically be identical except that it would use '1' as the upper limit and 2 as the multiplier (as opposed to '9' and 10).
That's the simplest form but there's plenty of scope for improvement to make your code more robust and readable:
Since the two functions are very similar, you could refactor out the common bits so as to reduce duplication.
You may want to consider an empty string as invalid rather than just returning zero as it currently does.
You probably want to detect overflow as an error.
With those in mind, it may be that the following is a more robust solution:
#include <stdbool.h>
#include <limits.h>
int convertBorCtoD(char *str, bool isBinary) {
// Configure stuff that depends on binary/decimal choice.
int maxDigit = isBinary ? '1' : '9';
int multiplier = maxDigit - minDigit + 1;
// Initialise accumulator to zero.
int retVal = 0;
// Optional check for empty string as error.
if (*str == '\0') return -1;
// Process each character.
while (*str != '\0') {
// Check character for validity.
if ((*str < '0') || (*str > maxDigit)) return -1;
// Add to accumulator, checking for overflow.
if (INT_MAX / multiplier < retVal) return -1;
retVal *= multiplier;
if (INT_MAX - (*str - '0') < retVal) return -1;
retVal += *str++ - '0';
}
return retVal;
}
int convertCtoD(char *str) { return convertBorCtoD(str, false); }
int convertBtoD(char *str) { return convertBorCtoD(str, true); }