recursive function: convert int to string

[shywolf91]

I need to create a recursive function that converts a int to a string.

The base case would be: input is from 0 to 9 and return the value as a string.

How would you return it a string?

I am not allowed to use atoi, itoa, for or while. Or anything similar to itoa or atoi.

It would look something like?

if(n==0) return 0?

[o3hC5Di1]

Hi there,

Have a look at the ASCII table values : http://en.cppreference.com/w/cpp/language/ascii.
You will notice that the decimal value 0 and the character 0 are 48 decimal numbers apart.

You can convert a character to an integer by abusing this fact as such:

char ch_n = '3';
int int_n = ch_n - '0';

Decimal character value of '3' in the ascii table is 51, we detract the decimal value of character '0' (48) from it - so we have the decimal number 3.

Hope that can help you get started.

All the best,
NwN

[shywolf91]

This is what I have so-far:

I need help with the base cases? I cannot use for or while loops.

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 (





}

[vlad from moscow]

Do you mean std::string or a character array?

If you mean a character array then the function is simple

#include <iostream>
#include <limits>

char * intToStr( char *p, unsigned int number )
{
	char digit = number % 10 + '0';

	if ( number /= 10 ) p = intToStr( p, number );

	*p++ = digit;

	return ( p );
}

int main()
{
   char s[std::numeric_limits<unsigned int>::digits10 + 2];

   unsigned int number = 123456789;

   char *p = intToStr( s, number );
   *p = '\0';

   std::cout << "s = " << s << std::endl;
}

[shywolf91]

I have something like this:

string intToStr(int n)
{
string str;
	if (n==0 || n==1 || ... ||n==9)
{
str=n;
return str;
}

}

[MetalMilitia]

Although my attempt isn't terribly readable, it's recursive and it handles both positive and negative integers correctly:

#include <iostream>
#include <string>
using namespace std;

string i_str(int num)
{
	string s;
	static bool recursive_call = false; //is this a recursive call?

	if(!recursive_call) {

		if(!num) return "0"; //deal with zero
		else if(num < 0) { //account for negative numbers
			num *= -1;
			s += '-';
		}
		recursive_call = true; //get ready for recursive calls
	}
	else if(!num) { //conversion is complete, reset function for future calls
		recursive_call = false;
		return "";
	}

	char app = num % 10 + '0'; //convert to a character
	return s + i_str(num / 10) + app; //append the characters in the correct order
}


int main()
{
	cout<< i_str(-456) << '\n' << i_str(-345) << '\n' << i_str(647)
		<< '\n' << i_str(235) << '\n' << i_str(4532175) << '\n' << i_str(-99567);
	cin.get();
}

[vlad from moscow]

If you need to use class std:string instead of a character array then the code could look the followimg way

#include <iostream>
#include <string>

std::string intToStr( unsigned int number )
{
	char digit = number % 10 + '0';

	return (  ( number /= 10 ) ? intToStr( number ) + digit : std::string( 1, digit ) );;
}

int main()
{
	std::cout << task1::intToStr( 123456789 ) << std::endl;
}