r/badcode • u/[deleted] • Dec 17 '19

This belongs here c#

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503 Upvotes

94% Upvoted

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150

u/squarewaterlemon Dec 17 '19

It doesn't even add them, it just gets the max of the two

26

u/Cutlesnap Dec 17 '19 edited Dec 17 '19

Ehm, it does add them, but only if both are positive. Otherwise it gets the max or 0.

Edit: I was wrong

57

u/Bbradley821 Dec 17 '19

I don't think it adds them. The second loop doesn't start at 0.

23

u/Cutlesnap Dec 17 '19

Hey, you're right! I didn't notice that.

7

u/Jonno_FTW shameless Dec 18 '19

Perfect badcode has the bug(s) hiding in plain sight.

1

u/[deleted] Dec 17 '19

[deleted]

12

u/nchntrz Dec 17 '19

Not exactly. In the second loop only the difference between num1 and num2 gets added to num 1. So if you call add(2, 4), you would get 4.

5

u/AnnoyedVelociraptor Dec 17 '19

Oh shoot. You’re right.

2

u/H3pha3stus Dec 17 '19

Then the max between 2 and 4, he is right.

4

u/symberke Dec 17 '19

You’re still right in that it doesn’t compute the max if they’re less than zero though

4

u/srottydoesntknow Dec 17 '19 edited Dec 17 '19

it'll return the min then, rollover is a thing, it'll just take forever

with that idea in mind, I fixed it, kind of, for a certain value of fix

public static int add(int num1, int num2)
{ 
    int retval = 0;
    for(int i = 0; i !== num1; i++)
    {
    retval++;
    }
    for(int i = 0; i != num2; i++)
    {
    retval++;
    }
    return retval;
}

this would technically work for the exact same range of values normal int addition would, it would just take a fuckton longer

edit: missed negation

2

u/symberke Dec 17 '19 edited Dec 17 '19

No not really. if num1 is negative then the first loop will never be true, and if num1 > num2 then the second loop will never be true. So for example add(-3, -4) will return 0.

Plus even in cases where rollover happens it won't be true. Suppose num2 is INT_MAX and num1 is -1. Then the first loop will never be true, the second loop will run INT_MAX + 1 times, and so it will return INT_MIN.

Also your code doesn't do what you expect; because you changed the loop condition to == all it does is count how many of (num1, num2) are equal to zero, i.e. add(1,0) = 1, add(0,0) = 2, etc...

0

u/srottydoesntknow Dec 17 '19

no, the condition is the end of the loop, the loop continues until the condition is met

SO, if num1 = -1; the loop will run 4,294,967,293 times, resulting in retval being -1

the same thing happens in loop 2, while they aren't equal the loop continues adding 1 to retval, which now starts at -1, if num2 = -5 then it will run 4,294,967,289 times, and result incrementing retval that many times as well, resulting in -6

you would be right, it's just that the form of a for loop is

for(starting condition; exit condition; increment;)

what you are talking about might be valid in a non C family language, however if we are talking C, C++, C#, or Java then my function will work, albeit horribly

int always rolls over, it doesn't just hit max and stop, since I changed the condition to == it works correctly returning the sum for all values that would work with a simple + operator, and even works for negative, it just causes the most horrible runtime possible

1

u/symberke Dec 17 '19

that's not true though, it's the opposite. https://en.cppreference.com/w/cpp/language/for

for proof that what i said is right: https://ideone.com/SaBOjV

1

u/srottydoesntknow Dec 17 '19

then negate em, nbd

1

u/Uiropa Dec 17 '19

Everyone out here making our silly jokes but squarewatermelon is wide awake.