Thread: itoa

  1. 12-11-2006 #1
    BeanieRaven
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    Smile itoa

    I have spent hours on this program and cannot get it to produce negatives or INT_MIN. Here is my code. Any help would be so appreciated I can't even begin to tell you...

    Code:
    #include <stdio.h>
#include <iostream>
using namespace std; 

char *strrev(char *);
char *itoa_3(int, char *, int);

int main()
 {
  char A[40]; 

  cout << "cout base 10" << endl; 
  cout << 17 << endl; 
  cout << 0 << endl; 
  cout << -17 << endl; 
  cout << INT_MAX << endl; 
  cout << INT_MIN << endl; 

  cout << "\ncout base 16" << endl; 
  cout << hex << 17 << endl; 
  cout << 0 << endl; 
  cout << -17 << endl; 
  cout << INT_MAX << endl; 
  cout << INT_MIN << endl; 
  cout << dec; 

  cout << "\nitoa_3 base 2" << endl; 
  cout << itoa_3(17, A, 2) << endl; 
  cout << itoa_3(0, A, 2) << endl; 
  cout << itoa_3(-17, A, 2) << endl; 
  cout << itoa_3(INT_MAX, A, 2) << endl; 
  cout << itoa_3(INT_MIN, A, 2) << endl; 

  cout << "\nitoa_3 base 10" << endl; 
  cout << itoa_3(17, A, 10) << endl; 
  cout << itoa_3(0, A, 10) << endl; 
  cout << itoa_3(-17, A, 10) << endl; 
  cout << itoa_3(INT_MAX, A, 10) << endl; 
  cout << itoa_3(INT_MIN, A, 10) << endl; 

  cout << "\nitoa_3 base 16" << endl; 
  cout << itoa_3(17, A, 16) << endl; 
  cout << itoa_3(0, A, 16) << endl; 
  cout << itoa_3(-17, A, 16) << endl; 
  cout << itoa_3(INT_MAX, A, 16) << endl; 
  cout << itoa_3(INT_MIN, A, 16) << endl; 

  cout << "\nitoa_3 base 32" << endl; 
  cout << itoa_3(17, A, 32) << endl; 
  cout << itoa_3(0, A, 32) << endl; 
  cout << itoa_3(-17, A, 32) << endl; 
  cout << itoa_3(INT_MAX, A, 32) << endl; 
  cout << itoa_3(INT_MIN, A, 32) << endl; 

  cout << "\nitoa_3 base 36" << endl; 
  cout << itoa_3(17, A, 36) << endl; 
  cout << itoa_3(0, A, 36) << endl; 
  cout << itoa_3(-17, A, 36) << endl; 
  cout << itoa_3(INT_MAX, A, 36) << endl; 
  cout << itoa_3(INT_MIN, A, 36) << endl; 

  return 0; 
} 

char *strrev(char *str) {
	char *p1, *p2;

	if (!str || !*str)
		return str;

	for (p1 = str, p2 = str + strlen(str) - 1; p2 > p1; ++p1, --p2) {
		*p1 ^= *p2;
		*p2 ^= *p1;
		*p1 ^= *p2;
	}

	return str;
}

char *itoa_3(int n, char *s, int b) {
	static char digits[] = "0123456789abcdefghijklmnopqrstuvwxyz";
	int i=0, sign;
    
	if ((sign = n) < 0)
		n = -n;

	do {
		s[i++] = digits[n % b];
	} while ((n /= b) != 0);

	if (sign < 0)
		s[i++] = '-';
	s[i] = '\0';

	return strrev(s);
}
  2. 12-11-2006 #2
    :davis:
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    Quote Originally Posted by BeanieRaven
    I have spent hours on this program and cannot get it to produce negatives or INT_MIN. Here is my code. Any help would be so appreciated I can't even begin to tell you...

    Code:
    #include <stdio.h>
#include <iostream>
using namespace std; 

char *strrev(char *);
char *itoa_3(int, char *, int);

int main()
 {
  char A[40]; 

  cout << "cout base 10" << endl; 
  cout << 17 << endl; 
  cout << 0 << endl; 
  cout << -17 << endl; 
  cout << INT_MAX << endl; 
  cout << INT_MIN << endl; 

  cout << "\ncout base 16" << endl; 
  cout << hex << 17 << endl; 
  cout << 0 << endl; 
  cout << -17 << endl; 
  cout << INT_MAX << endl; 
  cout << INT_MIN << endl; 
  cout << dec; 

  cout << "\nitoa_3 base 2" << endl; 
  cout << itoa_3(17, A, 2) << endl; 
  cout << itoa_3(0, A, 2) << endl; 
  cout << itoa_3(-17, A, 2) << endl; 
  cout << itoa_3(INT_MAX, A, 2) << endl; 
  cout << itoa_3(INT_MIN, A, 2) << endl; 

  cout << "\nitoa_3 base 10" << endl; 
  cout << itoa_3(17, A, 10) << endl; 
  cout << itoa_3(0, A, 10) << endl; 
  cout << itoa_3(-17, A, 10) << endl; 
  cout << itoa_3(INT_MAX, A, 10) << endl; 
  cout << itoa_3(INT_MIN, A, 10) << endl; 

  cout << "\nitoa_3 base 16" << endl; 
  cout << itoa_3(17, A, 16) << endl; 
  cout << itoa_3(0, A, 16) << endl; 
  cout << itoa_3(-17, A, 16) << endl; 
  cout << itoa_3(INT_MAX, A, 16) << endl; 
  cout << itoa_3(INT_MIN, A, 16) << endl; 

  cout << "\nitoa_3 base 32" << endl; 
  cout << itoa_3(17, A, 32) << endl; 
  cout << itoa_3(0, A, 32) << endl; 
  cout << itoa_3(-17, A, 32) << endl; 
  cout << itoa_3(INT_MAX, A, 32) << endl; 
  cout << itoa_3(INT_MIN, A, 32) << endl; 

  cout << "\nitoa_3 base 36" << endl; 
  cout << itoa_3(17, A, 36) << endl; 
  cout << itoa_3(0, A, 36) << endl; 
  cout << itoa_3(-17, A, 36) << endl; 
  cout << itoa_3(INT_MAX, A, 36) << endl; 
  cout << itoa_3(INT_MIN, A, 36) << endl; 

  return 0; 
} 

char *strrev(char *str) {
	char *p1, *p2;

	if (!str || !*str)
		return str;

	for (p1 = str, p2 = str + strlen(str) - 1; p2 > p1; ++p1, --p2) {
		*p1 ^= *p2;
		*p2 ^= *p1;
		*p1 ^= *p2;
	}

	return str;
}

char *itoa_3(int n, char *s, int b) {
	static char digits[] = "0123456789abcdefghijklmnopqrstuvwxyz";
	int i=0, sign;
    
	if ((sign = n) < 0)
		n = -n;

	do {
		s[i++] = digits[n % b];
	} while ((n /= b) != 0);

	if (sign < 0)
		s[i++] = '-';
	s[i] = '\0';

	return strrev(s);
}
    WTF are you trying to accomplish?

    Code:
    #include <stdio.h>
#include <string.h> /* for memset */

void swap( char* i, char* j )
{
    char c = *i;
    *i = *j;
    *j = c;
}

void reverse( char* str )
{
    char* i;
    char* j;
    for( j = str; *(j + 1); j++ );  /* get j to point at the last char in the str */
    for( i = str; i < j; i++, j-- ) /* swap i (first++) with j (last--) */
    {
	swap( i, j );
    }
}

void itoa( int n, char* str )
{
    int i;
    int sign;

    if( (sign = n) < 0 ) /* is n positive or negative */
    {
	n = -n; /* reverse the sign so that we work with a positive integer */
    }
    i = 0;
    do
    {
	str[i++] = n % 10 + '0'; /* add the ascii offset to the digit value */
    } while( (n /= 10) > 0 );
    if( sign < 0 )
    {
        str[i++] = '-'; /* put the sign back on */
    }
    reverse( str );  /* put the string in proper reading order */
}

int main()
{
    int i = 14593;
    int j = -84930;
    char buffer[BUFSIZ] = {0};

    itoa( i, buffer );
    printf( "%s\n", buffer );
    memset( buffer, 0, sizeof( buffer ) / sizeof( char ) );
    
    itoa( j, buffer );
    printf( "%s\n", buffer );
    
    return 0;
}

    :davis:
  3. 12-11-2006 #3
    BeanieRaven
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    I am trying to get output that looks like this:

    cout base 10
    17
    0
    -17
    2147483647
    -2147483648

    cout base 16
    11
    0
    ffffffef
    7fffffff
    80000000

    itoa_3 base 2
    10001
    0
    11111111111111111111111111101111
    1111111111111111111111111111111
    10000000000000000000000000000000

    itoa_3 base 10
    17
    0
    -17
    2147483647
    -2147483648

    itoa_3 base 16
    11
    0
    ffffffef
    7fffffff
    80000000

    itoa_3 base 32
    h
    0
    3vvvvvf
    1vvvvvv
    2000000

    itoa_3 base 36
    h
    0
    1z141yn
    zik0zj
    zik0zk
  4. 12-11-2006 #4
    Dave_Sinkula
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    Something like this may work:
    Code:
    #include <iostream>
#include <climits>
#include <cstdlib>
using namespace std;

int main()
{
   int size[] = {2,8,10,16,32,36};
   for ( size_t j = 0; j < sizeof size / sizeof *size; ++j)
   {
      const int array[] = {17,0,-17,INT_MAX,INT_MIN};
      char buffer[40];
      cout << "itoa base " << size[j] << ":\n"; 
      for ( size_t i = 0; i < sizeof array / sizeof *array; ++i )
      {
         cout << itoa(array[i], buffer, size[j]) << '\n';
      }
   }

   return 0;
}
    But itoa is non-standard.

    Or are you trying to prove out your itoa?
    7. It is easier to write an incorrect program than understand a correct one.
    40. There are two ways to write error-free programs; only the third one works.*
  5. 12-12-2006 #5
    BeanieRaven
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    Thanks Dave. I am trying to prove my version of itoa, but yours is producing the results I'm looking. I am sure to learn from it and solve my problem. Thank you again!
  6. 12-12-2006 #6
    Dave_Sinkula
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    Then this thread may be of interest.
    7. It is easier to write an incorrect program than understand a correct one.
    40. There are two ways to write error-free programs; only the third one works.*
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