Same algorithm implementation has different retults in Python and C++?

Question

This python code works properly and produces proper output:

def fib(x):
    v = 1
    u = 0
    for x in xrange(1,x+1):
        t = u + v
        u = v
        v = t
    return v

But when I write the same code in C++ it gives me a different and impossible result.

int fib(int x)
{
    int v = 1;
    int u = 0;
    int t;
    for (int i = 1; i != x + 1; i++)
    {
        t = u + v;
        u = v;
        v = t;
    }
    return v;
}

I'm still learning c++. Thanks!

Edit: C++ outputs -1408458269.
Python outputs 20365011074 when x = 50.

See Danial Fisher's answer below. In c++ types have specific sizes - you're overflowing a signed 32 bit int. — Brian Roach
– Brian Roach, Commented Jan 11, 2012 at 23:30

Daniel Fischer · Accepted Answer · 2012-01-11 23:24:44Z

16

For what input? Python has integers of unlimited (memory limited) size, C++'s int usually is a four byte integer, so you'll likely have overflow.

The largest Fibonacci number representable with a signed 32-bit integer type is fib(46) = 1836311903.

answered Jan 11, 2012 at 23:24

Daniel Fischer

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5 Comments

user875182 Over a year ago

Thank you. I used Unsigned int rather than int and it works perfect.

Mooing Duck Over a year ago

C++'s unsigned long long can go higher, up to fib(89). Above that you need a library.

Daniel Fischer Over a year ago

Not very far, the last you can get with 32 bits is fib(47), with 64 bits, it's fib(93).

Seth Carnegie Over a year ago

@Nekew you can also use unsigned long long for even bigger numbers (I think somewhere around 18446744073709551615 if your compiler makes it 8 bytes). Also, please click the checkmark beside this answer if it answered your question.

user875182 Over a year ago

I don't need to go larger than fib(50). Thank you for your help!

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