HI How do you convert a unsigned long to a string. Is there a function that i have overlooked or what. Thanks

Hello.

I actually anwered an almost identical question a few days ago. The ANSI C way to do this is to use sprintf from stdio.h. Just do this:

#include <stdio.h> // To use sprintf...unsigned long int x;x = 5;char *temp[2];sprintf(temp, "%d\0", x);printf("%s", temp); 

You could also use the plain ASCII set and do some calculations but it seems to be just harder work for the exact same result.

Minion

  ...int num = 37;char str = char(num)  

If you can''t do that in string class (string.h) (if they haven''t defined a string operator long()(), then you''re out of luck with that, unless you define your own (if you know how))

You could do it with sprintf, but it''s probably better to use itoa.

char Str[10];
itoa(Number, Str, 10);

The itoa function is not ANSI. It''s easy to write your own though.

[Resist Windows XP''s Invasive Production Activation Technology!]

hmmm... you''re right. If you need portability, write your own itoa clone

...using sprintf internally.

That couldn''t possibly work, you know... :D

you could always do something like this using remainders...

char * getNumberString(unsigned long * pDigitSpan, long number) {
/*
* get the number of digits in the number you''re converting
*/
unsigned long digitSpan = 0;

long tempNum = number;

while (tempNum > 0) {
tempNum /= 10;
digitSpan++;
}

/*
* actually build up the string of number
*/
char * numberString = new char[digitSpan];

tempNum = number;

for (unsigned long i = digitSpan; i > 0; i--) {
//see note below about this line
numberString[i-1] = tempNum % 10;

tempNum /= 10;
}

//i know you could probably incriment the *pDigitSpan
//instead of doing this assignment down here
//but i ran into some wierd problems doing something similar
//in another program so...
*pDigitSpan = digitSpan;


return numberString;
}

this also gives you the digit length of the number.

i think you need to convert tempNum % 10 from a single digit number into a char. i don''t remember exactly how to do that but i think it might be:

numberString[i-1] = ((char) tempNum % 10) - (char) 0;

damn, been way too long since i converted numerals into chars...

anyway, you can also use this function to convert to a binary number if you change the 10 to a 2 in all the % and /= statements. or even better would be to generecize the function by adding a passed int of the base number system you want to convert to. then use that instead of /= 10; (i.e. /= base

whatever, i know this function works for converting to binary. cause i wrote it for that purpose cept used a return type of bool * to conserve memory space.

-me

...that does not work if number == 0, or if you want to calculate the number in hex. Try this:

  const char char_values[] = "0123456789ABCDEF";char *IntToChar(int num, int base) {    int length = 0;    int temp_num = num;    do { // while(temp_num > 0)        ++length;        temp_num /= base;    } while(temp_num > 0);    char *str = (char*)malloc(length + 1);    str += length + 1;    *str = ''\0'';    do { // while(num > 0)        --str;        *str = char_values[num % base];        num /= base;    } while(num > 0);    return str;}