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itoa function (or similar) in Linux

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Join Date: Mar 2008

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marcosjp

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itoa function (or similar) in Linux

Sep 28th, 2008

Hello there!

How can I convert an INT value into a string?
I used the itoa function successfully in Windows:

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 c Syntax (Toggle Plain Text) 
itoa(the_int_number, the_string, 10);
itoa(the_int_number, the_string, 10);

It worked fine in Windows (DevC++) but Linux did not recognize this function.

Does anyone know another way of converting INT into char[n] in C that Linux could accept?

Thank you for any hints!

Marcos

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ssharish2005 ssharish2005 is offline

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Re: itoa function (or similar) in Linux

Sep 28th, 2008

Use sprintf function to convert int to string. Thats the more portable way of doing it. Since itoa is not portable!

 Help with Code Tags 
 c Syntax (Toggle Plain Text) 
char *cvtInt( char *str, int num)
{
    sprintf( str, "%d", num );
}
char *cvtInt( char *str, int num)
{
    sprintf( str, "%d", num );
}

ssharish

"Any fool can know, point is to understand"

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Join Date: Jul 2008

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ArkM

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Re: itoa function (or similar) in Linux

Sep 29th, 2008

The sprintf does not implement itoa functionality (radix in 2..36 range).
Fortunately, it's so simple to get itoa portable solution:

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 c Syntax (Toggle Plain Text) 
/* The Itoa code is in the puiblic domain */
char* Itoa(int value, char* str, int radix) {
    static char dig[] =
        "0123456789"
        "abcdefghijklmnopqrstuvwxyz";
    int n = 0, neg = 0;
    unsigned int v;
    char* p, *q;
    char c;
 
    if (radix == 10 && value < 0) {
        value = -value;
        neg = 1;
    }
    v = value;
    do {
        str[n++] = dig[v%radix];
        v /= radix;
    } while (v);
    if (neg)
        str[n++] = '-';
    str[n] = '\0';
    for (p = str, q = p + n/2; p != q; ++p, --q)
        c = *p, *p = *q, *q = c;
    return str;
}
/* The Itoa code is in the puiblic domain */
char* Itoa(int value, char* str, int radix) {
    static char dig[] =
        "0123456789"
        "abcdefghijklmnopqrstuvwxyz";
    int n = 0, neg = 0;
    unsigned int v;
    char* p, *q;
    char c;

    if (radix == 10 && value < 0) {
        value = -value;
        neg = 1;
    }
    v = value;
    do {
        str[n++] = dig[v%radix];
        v /= radix;
    } while (v);
    if (neg)
        str[n++] = '-';
    str[n] = '\0';
    for (p = str, q = p + n/2; p != q; ++p, --q)
        c = *p, *p = *q, *q = c;
    return str;
}

Last edited by ArkM; Sep 29th, 2008 at 3:03 am.

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sladevi

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Re: itoa function (or similar) in Linux

Oct 21st, 2008

That's handy ArkM, but could you provide some references? How do you know the code is in the public domain and where did you get it from?

Also, is it just me or is that code bugged? When n/2 is odd, the for loop continues infinitely as p and q simply bypass each other, so the condition should actually be p<q, right?

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ArkM

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Re: itoa function (or similar) in Linux

Oct 21st, 2008

Ohh, that's mea culpa... See right code below...
I know that this code is in the public domain because I have wrote it for 5 (or 10

) minutes 22 days ago. So I have got it directly from my VC++ 2008 solution...

 Help with Code Tags 
 c Syntax (Toggle Plain Text) 
/* The Itoa code is in the public domain */
char* Itoa(int value, char* str, int radix) {
    static char dig[] =
        "0123456789"
        "abcdefghijklmnopqrstuvwxyz";
    int n = 0, neg = 0;
    unsigned int v;
    char* p, *q;
    char c;
 
    if (radix == 10 && value < 0) {
        value = -value;
        neg = 1;
    }
    v = value;
    do {
        str[n++] = dig[v%radix];
        v /= radix;
    } while (v);
    if (neg)
        str[n++] = '-';
    str[n] = '\0';
 
    for (p = str, q = p + (n-1); p < q; ++p, --q)
        c = *p, *p = *q, *q = c;
    return str;
}
/* The Itoa code is in the public domain */
char* Itoa(int value, char* str, int radix) {
    static char dig[] =
        "0123456789"
        "abcdefghijklmnopqrstuvwxyz";
    int n = 0, neg = 0;
    unsigned int v;
    char* p, *q;
    char c;

    if (radix == 10 && value < 0) {
        value = -value;
        neg = 1;
    }
    v = value;
    do {
        str[n++] = dig[v%radix];
        v /= radix;
    } while (v);
    if (neg)
        str[n++] = '-';
    str[n] = '\0';

    for (p = str, q = p + (n-1); p < q; ++p, --q)
        c = *p, *p = *q, *q = c;
    return str;
}