actually:
log10(x) = ln(x)/ln(10)
i thought the standard log() function was base2? not sure.
===============================================
If there is a witness to my little life,
To my tiny throes and struggles,
He sees a fool;
And it is not fine for gods to menace fools.
int 123 to array[3] = {1, 2, 3}
Why wouldn''t you use the itoa() function? Int-To-Array... Am I missing something here?
Regards,
Jumpster
Regards,
Jumpster
Selkrank, Mithrandir: Officially log is indeed 10 based (and yes, log(x) = ln(x)/ln(10), I stand corrected), but sometimes log is 2 based or even log == ln, especially in software. I guess a guy might get confused 
.
Selkrank: floor() just chops of the decimals from a double (e.g. floor(0.5) == 0.0). I guess you could rely on the compiler''s implicit type casting abilities instead.
Of course you can use itoa, but then you''ll get a char array, not an int array, and you would have to subtract ''0'' from each char to get its actual value. I think it''s faster to just do it yourself, because who knows what that itoa function does inside?
Dormeur
.Selkrank: floor() just chops of the decimals from a double (e.g. floor(0.5) == 0.0). I guess you could rely on the compiler''s implicit type casting abilities instead.
Of course you can use itoa, but then you''ll get a char array, not an int array, and you would have to subtract ''0'' from each char to get its actual value. I think it''s faster to just do it yourself, because who knows what that itoa function does inside?
Dormeur
Here. No expensive math routines needed:
int num=12345;
int digits[MAX_DIGITS];
char buf[MAX_DIGITS];
int i;
itoa(num, buf, 0); // or sprintf() if you like
for(i=0; i digits = (int)buf - 0x30;<br><br> </i>
int num=12345;
int digits[MAX_DIGITS];
char buf[MAX_DIGITS];
int i;
itoa(num, buf, 0); // or sprintf() if you like
for(i=0; i digits = (int)buf - 0x30;<br><br> </i>
That last one got messed up.
int num=12345;
int digits[MAX_DIGITS];
char buf[MAX_DIGITS];
int i;
itoa(num, buf, 0); // or sprintf() if you like
for(i=0; i(lessthan)strlen(buf); i++)
digits = (int)buf - 0x30;<br><br> </i>
int num=12345;
int digits[MAX_DIGITS];
char buf[MAX_DIGITS];
int i;
itoa(num, buf, 0); // or sprintf() if you like
for(i=0; i(lessthan)strlen(buf); i++)
digits = (int)buf - 0x30;<br><br> </i>
is it me??
nahh it''s you
welll
int x = 13464, i = 0;
while(x > 0)
{
arr[i++] = x % 10;
x /= 10;
}
so now you don''t have to use STRING funcs
hope it helps good luck
http://qsoft.cjb.net
nahh it''s you
welll
int x = 13464, i = 0;
while(x > 0)
{
arr[i++] = x % 10;
x /= 10;
}
so now you don''t have to use STRING funcs
hope it helps good luck
http://qsoft.cjb.net
Guys, you''re making this _way_ too complicated.
int whee = 12345;
char str[32];
sprintf(str, "%d", whee);
_strrev(str);
If your compiler doesn''t have strrev(), write your own - its a trivial function.
int whee = 12345;
char str[32];
sprintf(str, "%d", whee);
_strrev(str);
If your compiler doesn''t have strrev(), write your own - its a trivial function.
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