x/y a runtime error

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3.00/5 (3 votes)

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I am using c# 4.0
I have an issue:

C#

int x=10;
int y =0;
int z=x/y;  //<-- this gives an error at runtime..

however,

C#

int z=10/0;  //<-- this gives an error at compile-time....

May I know why this is the case?

Thanks & Regards,

Posted 11-Jul-11 1:36am

shikhar gilhotra

Updated 11-Jul-11 1:49am

Manfred Rudolf Bihy

v2

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Emrah Taylan 11-Jul-11 7:44am

Because on compile action your ide dont look y declaration is 0 for this reason it gives on runtime

but second example this 0 for this reason it will sow on compile action!

ok?

4 solutions

Solution 1

Hi,

Value in x and y variable setting at run time. Where in 10/0 both are constant, their value setting at compile time.

Posted 11-Jul-11 1:47am

vivekse

Solution 3

Side note to the other answers:
This is dependent how far the parser is "looking" and preprocessing ahead. If you would use Delphi for example, both would have given you a compile time error because it does some serious preprocessing.

Good luck!

Posted 11-Jul-11 1:57am

E.F. Nijboer

Comments

Manfred Rudolf Bihy 11-Jul-11 8:02am

As I thouroghly explained in my answer, detecting the first case is not an issue of preprocessing (unless improperly labled thusly by the compiler manufacturer), but rather correctly inferring what the variables values will be at runtime. This can be "guessed" (inferred if you prefer) best for expression sets where definitions and declarations coincide in the same statment.
Parser lookahead also doesn't have the least to do with that.

Sergey Alexandrovich Kryukov 11-Jul-11 21:47pm

Interesting. I used Delphi from its day zero (and years more of Borland Pascal), but don't remember that. My 5.
I am somewhat doubtful it makes sense -- please see my solution.
--SA

Solution 4

A note to the correct answer by Manfred:

Run-time behavior is not inferred by a good reason: in general case this is impossible. This is the consequence of one of fundamental theorems of computability. If it is not possible based on static analysis in general case, why detecting some cases? Just to create false illusion of safety in developers (and thus provoking errors)?

By that reason, no attempt to statically analyze code to find such problem is done — and this is very good.

As to the expressions like 8/2, the whole expression is treated as one single immediate constant, evaluated during run-time and put in the byte code as one constant, not as expression. Naturally, integer division by zero fails. :-)

A note: pay attention that division by floating-point zero is quite legal and produce valid value of Inf. You can continue using the result; for example, 1/Inf will be evaluated as 0.

—SA

Posted 11-Jul-11 15:44pm

Sergey Alexandrovich Kryukov

Comments

Espen Harlinn 12-Jul-11 9:56am

Good points - my 5

Sergey Alexandrovich Kryukov 12-Jul-11 10:36am

Thank you, Espen.
--SA

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Manfred Rudolf Bihy · Accepted Answer · 2011-07-11T01:47:00

That is because the first error can be only be reliably detected at runtime. The second error is detected at compile time because the compiler could already execute the division when compiling the program. Like when this is done for optimization purposes. The first case poses a problem for the compiler because it is hard/impossible to infer the correct runtime condition just from the given code. Actually in your case where the declaration coincides with the definition the compile should be able to detect this case also. I think the real reason why this is not checked is that this is a very trivial case where runtime behaviour could be infered, whereas it would be impossible in most all of the other possibilities.

Cheers!

—MRB

x/y a runtime error

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