log10(8)

[Jay B. Harlow]

In article <3244 at servax0.essex.ac.uk> elzea at sersun0.essex.ac.uk (El-Zein A A) writes:
>
>	double l;
>	.
>	l = log10(8);
		  ^  this is a integer!
>	printf("%f", l);
>

simple, (is this a trick question? ;-))

because the 8 you sent as an argument to log10 is a integer, if you did not
include math.h (where log10 is declared) log10 would return a '0.00000' 
(mine does ;-)  Any way because your compiler does not have a PROTOTYPE 
(YEA ANSI C) it is not smart enough to convert the integer 8 to
a double (8.0) which is what log10 expects. 
You do realize that a integer 8 is almost never the same binary format
(as an argument to a function) as a double is? so log10 was
reading extra stack stuff as a floating point number.

			Jay
-- 
		Jay B. Harlow	<harlow at plains.nodak.edu>
	uunet!plains!harlow (UUCP)	harlow at plains (Bitnet)

Of course the above is personal opinion, And has no bearing on reality...