segmentation fault in pi calculation (python)

Question

def pi(times):
    seq = []
    counter = 0
    for x in range(times):
        counter += 2
        seq.append("((%f**2)/(%f*%f))*"%(float(counter), float(counter-1), float(counter+1)))
    seq.append("1.0")
    seq = "".join(seq)
    seq = eval(seq)
    return seq*2

Anywhere past 85000 terms I get a segmentation fault and python quits. How can I avoid this? Why is it crashing? Can't it just please use more memory or something?

Why are you doing the calculation that way, instead of just doing a running calculation as you figure out the terms? Also, why are you reusing `seq` for the joined string and the evaluated result? — ncoghlan, Mar 03 '11 at 06:59

score 5 · Accepted Answer · answered Mar 03 '11 at 07:02

5

You appear to have found a bug in eval where it can't handle insanely long expressions:

>>> eval("1.0*"*10000+"1.0")
1.0
>>> eval("1.0*"*100000+"1.0")
# segfault here

I use the phrase "insanely long" advisedly though. Don't do it that way, calculate the pieces as you go. There is no reason to be using eval in this situation.

answered Mar 03 '11 at 07:02

ncoghlan

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wow thank you :) now I'm getting a syntaxerror when trying to return seq as an int, calculating on the fly – tekknolagi Mar 03 '11 at 07:05
i cannot believe i didn't calculate on the fly instead of using eval :O – tekknolagi Mar 03 '11 at 07:07
I've seen something about this before and it isn't really considered a bug so much as the programmer abusing eval. – Justin Peel Mar 03 '11 at 07:30
5

The segfault outcome is a bug. Simply declaring an expression that long to be unreasonable and refusing to accept it would be OK, but it should raise an exception in that case. – ncoghlan Mar 03 '11 at 07:58
Ah, it was a known defect, it just didn't have a tracker item. – ncoghlan Mar 03 '11 at 11:11

score 2 · Answer 2 · answered Mar 03 '11 at 07:06

2

Why use eval() at all?

def pi(times):
    val = 1
    counter = 0
    for x in range(times) :
        counter += 2
        val *= float(counter)**2/(counter**2 - 1)
    return val * 2

Does the exact same thing.

answered Mar 03 '11 at 07:06

NullUserException

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segmentation fault in pi calculation (python)

2 Answers2

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