pow() returns Infinity?
Hi
I'm trying to evaluate 2^19937 with the pow function.
In this case I'm using the following code:
However res is evaluated as 1.#INF000000000000, what am I missing here?
Many thanks
I'm trying to evaluate 2^19937 with the pow function.
In this case I'm using the following code:
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However res is evaluated as 1.#INF000000000000, what am I missing here?
Many thanks
2^19937 is a pretty massive number... maybe TOO long to store in a var, try this...
std::cout << pow(b, n);
2^19937 is a GIANT number...over 6000 digits.
There's no way you could possibly fit that in any of the fundamental data types.
In fact, it's (sorry for wall of numbers)
What do you need such a big number for?
There's no way you could possibly fit that in any of the fundamental data types.
In fact, it's (sorry for wall of numbers)
43154247973881626480552355163379198390539350432267115051652505414033306801376580 91130451362931858466554526993825764883531790221733458441390952826915460916801900 78753437413962968019201144864809026614143184432769803000667281049840954515881760 77132969843762134621790396391341285205627619600513106646376648615994236675486537 48024196435029593516866236390904794834769231397830137782078571241905447433284452 91831729732423108882650813216264694510777078122828294447750226804880578200287646 59399164766265200900561495800344054353690389862894061792872011120833614808447482 91354732836727787956564830784690911694586623016970240126024018702874665003344577 45703154312929960251877807901193759028631710841496424733789862675033089613749057 66340905289572290016038000571630875191373979555047468154333253474991046248132504 51634179655147057548145920085947261483621387555711686444578975088627799648730430 84504842234206292665185560243393391908443689210184248446770427276646018529149252 77280922697538426770257333928954401205465895610347658855386633902546289962132643 28242574803578623358060815469654693256383332767076989943977488852668727852745100 29630591469638757154257355344759797344631006783673933274021499309687782967413915 14599602374213629898720611431410402147238998090962818915890645693934483330994169 63229587799584899336674701487176349480554999616305154122540346529700772114623135 57040814930986630657336771911728539870957481678162560842128233801686253345864312 54034670806135273543270714478876861861983320777280644806691125713197262581763151 31359642954776357636783701934983517846214429496075719091805462511414366638418943 38525764522893476524546315357404687862289458856546085620580424689873724369214450 92315377698407168198376538237748614196207041548106379365123192817999006621766467 16711347163271548179587700538269439340040306170045769113534918787488892342934934 01451705717161811257958888892774954269771499145496239163940148229850253316515114 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What do you need such a big number for?
yea i am, thank you!
so taking into account practical considerations, how would you calculate the reciprocal (1/(1+2^19937)) if you can't readily store such a massive number in memory?
so taking into account practical considerations, how would you calculate the reciprocal (1/(1+2^19937)) if you can't readily store such a massive number in memory?
Last edited on
That number...would be very close to 0.
I don't know how accurate this is, but it could possibly be approximately equal to 2.317 * 10^-6002.
You might want to look into arbitrary precision number libraries. (that's what I used to calculate 2^19937)
Such as: http://gmplib.org/
I don't know how accurate this is, but it could possibly be approximately equal to 2.317 * 10^-6002.
You might want to look into arbitrary precision number libraries. (that's what I used to calculate 2^19937)
Such as: http://gmplib.org/
Last edited on
yea, adding another level of detail, this was all part of building a random variate generator with boost. so the idea of calculating the reciprocal was to convert the mersenne twister generated number to a uniform variate using the existing structure.
it's a bit of a learning experience right now so i've opted to go with the boost variate generator to just get the value out.
thanks all again for your help.
it's a bit of a learning experience right now so i've opted to go with the boost variate generator to just get the value out.
thanks all again for your help.
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