Convert to ASCII ?

[Vaclav_]

This seems very silly and basic - The function description is pretty clear Convert an unsigned long integer into a string, using a given base So why I keep printing numbers and NOT ASCII letter? I need a pointer to string - not numeric value. And I am sorry to ask such stupid question. Cheers Vaclav

int base = 10;
long value = 73;
char buffer[33];
unsigned long RANDOM = random(65, 79);

for( base = 2; base <= 16; base = base + 2 ) {
    printf( "%2d %s\\n", base,
        ultoa( RANDOM, buffer, base ) );
}

[CPallini]

I suppose, you know that

printf("%d\n", ul);

and

printf("%s\n", ultoa(ul, buffer, 10));

should produce the same output.

[Lost User]

Vaclav_Sal wrote:

So why I keep printing numbers

Because that is what ultoa creates. It takes a numeric value (long) and converts it to printable characters. Try an input value of 127 instead of RANDOM.

[Vaclav_]

So itoa is really NOT "integer to ASCII" conversion, right? I knew it was a silly question. I suppose I can use some kind of lookup method the get from random index # to character array of alphabet. I'll check stdio. Thanks

[Richard Andrew x64]

Try this:

 for( base = 2; base <= 16; base = base + 2 ) {
        ultoa( RANDOM, buffer, base );
        printf( "%2d %s\\n", base, buffer );
 }

The difficult we do right away... ...the impossible takes slightly longer.

[Lost User]

Vaclav_Sal wrote:

So itoa is really NOT "integer to ASCII" conversion, right?

Yes, of course it is. It converts a binary integer to a set of printable (ASCII) characters. What else would you expect it to do?

[CPallini]

If all you need is the ASCII character corresponding to the given integer than a cast is enough, e.g.

printf("%c\n", (char) RANDOM);