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let's use some impossibly high numbers
57190723512387946763891783751834690731790415789487683174589713456791364901495019099013980475738967773571904779137485784758475667135607897897897897894578+e82940580290183467317613189704757891390570918923785899018394*589000000071023485810845780384581023809048108289359090242315081238949209385481390475012094890892390905890810239458+e082807509890578890787819384579=

Even though it's an exponential, there's nothing to make sense out of it. At least a 420³ or 666² would make sense. But as it is, it's nothing but a mash of keyboard slam numbers xP

THE WALL OF MATH HAS AWOKEN!
76190893478561789982347590292+929387578990103795910024578923*5990901283859\280590\2275687e758290909359293945894*2*2*2*2*2*88854770020387540912938e99273478039540\\\\\\\\\\\\\\\\\\\\\\\\\\74822222222222222222546222222222224222226782379379379379379379379379379379379379379379379379379379379379379379379379379379379379379379379379379379379379379379379379379379e189189189189189189189189189189189

BUT WE'RE GONNA FIGHT IT, I'M EQUIPPED WITH ANTIMATH SWORD THAT DOES 57020375902983905802993890590289890579081750+e99457802893450 DAMAGE TO MATH MONSTERS

Theo won!
Theo got 9402193549019034+e9928945 XP
Theo leveled up!
Theo leveled up!
Theo leveled up!
Theo leveled up!
Theo leveled up!
Theo leveled up!
Theo leveled up!
Theo leveled up!
Theo leveled up!
Theo leveled up!

I'll solve it for ya.
(x + 1)^2 + x^2 + x = 0
(x + 1)^2 + x(x + 1) = 0
(x + 1)(x + 1 + x) = 0
(x + 1)(2x + 1) = 0
So it should be:
x + 1 = 0 or 2x + 1 = 0
x = -1 or 2x = -1
x = -1 or x = -1/2 = -0,5
In conclusion, x equals either minus one or minus zero point five. Try replacing x with either.

I wouldn't call it hard. However I wouldn't call it easy when it's worded in this way, as it's pretty hard to organize an equation like that on a profile post. Mathematics, while not being my forte, is a subject I like and appreciate learning about.

what about this
e+at88920- 629g8ujhje8t*fuj2004tjjnweoiert/g88932jhjhjgooskejfo23489t9923uhtfgl+e8849921fjnnsiignerioujgbutts95230549'102313371337PINGAS

anyone?
ok i'll just throw some more
8189357891389450134789057901890234789051708934905692378978968991347893478568345885981739847671819678989478956089348957931467318945798845897674189489584179589234809578901356873891896913048573895673789056713967138906738968138986930478963896889576767798031057634067354684568768389589417905679314858745843910475588901783489579019084578+e901848904903010=

@MarioKart7z That's not a valid for-loop, you set i to 0 and check if it's greater than 97589273489057802909859029093+e99547220, which is far greater than the maximum capacity of an unsigned int, and the loop doesn't run ಠ_ಠ

They can, but you tend to get the typical floating point problems. As an example, incrementing by 0.1f each iteration will get you 7.79999 instead of 7.8 in some places.

I'm pretty sure the reason that happens is because of the amplification of the smallest value possible (denoted as Epsilon in both float and double in System.dll)

Yep, that also generally happens because fp's use fixed amount of binary numbers to represent the decimal number (remember, a standard floating point only has 4 bytes of precision), so it can't represent numbers accurately. My favorite example is setting a float to 1/3 and printing out the result.

## Comments on Profile Post by MarioKart7z

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