CODEWITHSUNDEEP
c++
ssrobbi
ssrobbi

Reputation: 571

How to convert int to string (in c++) without access to int to string function

So I got asked this on a skills test for an internship, and at the time it completely baffled me. I have some code now that I thought should work, but isn't assigning the correct values to the string.

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

int main()
{
     // declarations
     int i = 0, num= 63;
     string b="";

     while (num != 0)
     {
          i = num % 10;
          b.insert(0,i + 48 + "");
          num = num/10;
     }


     cout << b << endl;

     return 0;
}

With "num" being the integer value (I just used 63 as an example)

EDIT:: I incorrectly paraphrased, I could not use any function that did an int to string conversion for me, not that I couldn't use the string library, my mistake.

Upvotes: 4

Views: 4439

Answers (5)

Kerrek SB
Kerrek SB

Reputation: 477020

Here's some code that doesn't use any libraries and doesn't perform allocations. Instead, the user is expected to provide a buffer that's large enough, and she receives a pointer to the beginning of the result string in return:

char * format(unsigned int n, char * buf, std::size_t len)
{
    if (len == 0) return NULL;

    char * p = buf + len;

    *--buf = 0;

    while (n && p > buf)
    {
        *--buf = '0' + (n % 10);
        n /= 10;
    }

    return n == 0 ? p : NULL;
}

Embedded people might like this especially because the function doesn't do anything outside your control.

Usage:

char buf[100];
char * s = format(12345, buf, sizeof buf);
if (s) { printf("Formatted: %s\n", s); }

Upvotes: 0

Casey
Casey

Reputation: 10936

Very simple solution: Use string streams. In this case, an output string stream:

#include <iostream>
#include <string>
#include <sstream>

std::string IntToString(int value) {
    std::ostringstream ss;
    ss << value;
    return ss.str();
}

const char* IntToCString(int value) {
    std::ostringstream ss;
    ss << value;
    return ss.str().c_str();
}

Upvotes: 0

Dietmar K&#252;hl
Dietmar K&#252;hl

Reputation: 153820

It is worth noting that there are other errors with the code. For example, the transformation doesn't work for num == 0. Here is a better version:

#include <iostream>
#include <algorithm>
#include <string>
#include <cstdlib>

int main(int ac, char* av[])
{
    int num(ac == 1? 0: std::atoi(av[1]));
    std::string b;

    do
    {
        b.push_back((num % 10) + '0');
        num /= 10;
    }
    while (num != 0);
    std::reverse(b.begin(), b.end());
    std::cout << b << '\n';
    return 0;
}

Upvotes: 0

Grzegorz
Grzegorz

Reputation: 3335

I would suggest recursive function:

std::string f( int val )
{
   if ( val >= 10 )
      return f( val / 10 ) + ( char )( val % 10 + '0' ) ;
   else
      return std::string( 1, '0' + val ) ;
}

Usage:

std::cout << f( 63 ) << std::endl ;

Note that this function should append to your string instead of using very expensive shifting (and putting in the front of the string.)

Upvotes: 0

Praetorian
Praetorian

Reputation: 109119

Change the insert line to 

b.insert(0, 1, '0' + i);

This will insert the character obtained from adding i to '0' once at index 0 in the string.

Explanation of why your code isn't working:

b.insert(0,i + 48 + "");

the line above is calling 

basic_string::insert( size_type index, const CharT* s );

where the pointer is being determined by adding 48 + i to the address of the "" string literal, resulting in garbage.

Upvotes: 6

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