[solved][C] input validation with strtod()

Question

I'm using strtod() in an RPN implementation, and I'm working on input validation. As per strtod(3), I've written the following conditional to catch input overflow.
errno = 0;
op1 = strtod(token, &endPtr);
if ((errno == ERANGE && (op1 == HUGE_VALF || op1 == HUGE_VALL)) || (errno != 0 && op1 == 0)) {
printf("Error: Input overflow.\n");
clearstack();
return 1;
Except that when I enter something that would definitely overflow, this conditional is never entered. Furthermore, errno is never set. What am I missing?
A link to the full code can be found here
Last edited by falconindy (2010-04-13 17:49:15)

Answer 1

tavianator wrote:
Well the first wrong thing I see is that you should really test if fabs(op1) is HUGE_VAL{F,L}.
The second thing is, why so complicated a test?  errno != 0 should be all you need.
Fair enough. This is the first time I'm really dealing with floating point ops in C.
tavianator wrote:The third thing is, if errno is really not being set, are you sure you're really overflowing?  doubles go up to about
1.8e308 on most arches.
Huh... wouldn't an unsigned double be, at most, 2^128 - 1 and more likely 2^64 - 1? I'm dealing with signed, so its then half that. Sure enough, sizeof(double) returns 8.
Also, this is what made me think I'm hitting overflow.
> 111111111111111111111111111111111111111111111111111111111111111111111111111111111111 1 *
= 111111111111111105547424803279259724863245197668615715838829352563602489955831513088.000
>
Hmm. If I go even further, I eventually do hit an overflow error. However, that still leaves me a little baffled as to the results above. Is this a result of the decimal precision inherent in a double?