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I already completed one part of the assignment, but need help on the other part. The assignment was to create two recursive functions. One that converts strings to integers ( which i completed) and the other converts integers to strings.
I need help on the last function.
This is the prototype: string intToStr(int n);
What have you tried? Can you use the ostringstream class?
Jim
If you are not allowed to use ostringstream class,which probably don't,you may have to take a look at itoa itoa - C++ Reference
We are not allowed to use any functions that are similar to atoi or itoa. We also cannot use for or while.
This is my other function that converts the string to an integer:
Code:int stringToInt(string str)
{
static bool OK; //This variable signals if the string can be treated as int
static int sign; //This variable signals about the sign of the number
OK = true;
sign = 0;
int digit = 0; // The result of the function
int length = str.length();
if (length == 2) {
if (str[0] == '-')
sign = -1;
else if (str[0] == '+')
sign = +1;
}
if (length == 1 || length == 2 && sign != 0) {
if (str[length-1] < '0' || str[length-1] > '9') {
OK = false;
return 0;
}
digit = str[length-1] - '0';
if (sign == -1) digit = -digit;
}
// The main case:
else {
digit = stringToInt(str.substr(0, length-1));
if (!OK) return 0;
digit *= 10;
if (str[length-1] < '0' || str[length-1] > '9') {
OK = false;
return 0;
}
if (sign == -1) // If the sign is minus, then do it by subtracting
digit -= str[length-1] - '0';
else // otherwise, do it by adding
digit += str[length-1] - '0';
}
// Return the final result
return digit;
}
That seems a bit large, so probably a lot that can be improved upon there. Are you sure that you're allowed to use static variables?
You still haven't posted your attempt at intToStr though, or asked any questions about it, so we can't help you with that yet. The good news is that it's easier than the other way around.
Not sure about static variables. As for intToStr. I haven't started it yet. I would like some help with the base case of the recursive part.Quote:
Originally Posted by iMalc
The base case would be that the input is from 0 to 9, and all you need to do is return that digit as a string.
Yes the base case really is that simple!
That clearly leaves negative values, and values of 10 or greater, as remaining cases. Now code that up and have a go at the remaining cases then post your attempt.
So it would be like this? if(n==0) return 0?Quote:
Originally Posted by iMalc
How would you return it as a string? Also could you also provide some advice on the main function? Would I be using the same main function as I did for stringToInt? Thank-you!
That obviously only works for zero, but I said the base case would handle 0 to 9.Quote:
Originally Posted by Dylan Metz
I would probably use the std::string constructor that takes a count and a character, to build a string of 1 character that results from adding character '0' to the digit.
Your main would obviously have to change because you'll be passing in an int instead of a string, and returning a string instead of an int. Not to mention you'll be calling a different function. Yes I'm really stating the obvious here, so if it doesn't answer the question, please ask a better question.Quote:
Also could you also provide some advice on the main function? Would I be using the same main function as I did for stringToInt?
I honestly expected that having written stringToInt, you would have no trouble writing intToString. How you managed to write the harder one and are then stuck on the easier one, I don't know.
Here is what I have so far:
I need help with setting up the base case. I think it looks like this for int 0?Code:#include <iostream>#include <string>
using namespace std;
//recursive function prototype
string intToStr(int n);
//main
int main()
{
/*/int tests[] = {
0, 100, -5, 10, 5, 10302
};
const int B = 6;
for (int i=0; i<B; i++)
{
cout << "Converting the integer "<< tests[i] <<" to string: ";
cout << intToStr(tests[i]) << endl;
}*/
int num;
cout<<"Please input a number: ";
cin>>num;
cout<<endl;
cout<<"Converting the integer "<<num<<" to a string: ";
cout<<intToStr(num)<<endl;
//string str;
system("PUASE");
return 0;
}
string intToStr(int n)
{
if (
}
if(?==0) return 0;
How do you convert that to a string and return it?
If ostringstream was not allowed yet itoa was allowed, I'll facepalm and then blast the instructor since ostringstream is standard and itoa is non-standard, yet the effect of getting a library routine to do the work is the same.Quote:
Originally Posted by std10093
It would be:Quote:
Originally Posted by Dylan Metz
The explicit conversion to string is optional though.Code:if (n == 0)
{
return string("0");
}
How would the remaining cases look like? I sorta remember my instuctor talking about using either division or modulo(%)?
If you wrote the other function then you can certainly do a lot more than just "if (", especially after all the help we have provided already.
It really starts to seem like you either copied the other function from someone else or you're really not trying very hard at all.
As much as it may sound unfair suggesting this, we definitely do get people playing dumb to let others do the work for them.
Unless you post a question about something we've said, or post an attempt with significantly more headway, I'm going to have to leave it to you to do some experimentation and learning on your own.
Please, have a good think about it, and look stuff up on the internet when you have a general idea but cant figure out the specifics.
Besides, you learn a lot more when not being spoon-fed.
Are you failing with the logic (ie, how does on accomplish such a face)? Then I'd suggest you make a flowchart to figure out the thinking.
If you are failing with some syntax or don't know what facilities exist to achieve something, then by all means, ask away.
I'd like to at least see that you have a way in mind on how to accomplish this behavior. Post your throughts.