Stupid compiler error

[rollasoc]

Hi,

Made me laugh so I though I'd share this.

short i = 32;
short j = 1;
short k = i + j;

gives an error
Cannot implicitly convert type 'int' to 'short'

That is a bit poo in my reckoning...

at least the following compiles.

short k = (short)(i + j);

[Morten Wennevik]

The same thing happens in Java, and there it seems that '+' will return
"at least int, because of normal promotions".
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[Malik Arykov]

Hello, rollasoc!

r> short i = 32;
r> short j = 1;
r> short k = i + j;

int k = i + j; // Ok

[Morten Wennevik]

You will get the same if you try

byte = byte + byte;

I wonder why

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[rollasoc]

Indeed true

but that won't work when I the call a C++ function expecting a short.
"Malik Arykov" <ar****@r40.nalog.ru> wrote in message
news:##**************@TK2MSFTNGP11.phx.gbl...

Hello, rollasoc!

r> short i = 32;
r> short j = 1;
r> short k = i + j;

int k = i + j; // Ok

[Morten Wennevik]

The same thing happens in Java, and there it seems that '+' will return
"at least int, because of normal promotions".
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Using M2, Opera's revolutionary e-mail client: http://www.opera.com/m2/

[Sherif ElMetainy]

Hello

The CLR can perform arithmetic operations on 32 bit or 64 bit values only.
It can't perform 16 bit or 8 bit operations.
So the i and j are implicitly converted to Int32, then added, so the result
of the addition is an Int32, which have to be explicitly cast to short to
get a short result.

Best regards,
Sherif

"rollasoc" <ho*****@hotmail.com> wrote in message
news:uz**************@TK2MSFTNGP10.phx.gbl...

Hi,

Made me laugh so I though I'd share this.

short i = 32;
short j = 1;
short k = i + j;

gives an error
Cannot implicitly convert type 'int' to 'short'

That is a bit poo in my reckoning...

at least the following compiles.

short k = (short)(i + j);

[Ignacio Machin \( .NET/ C# MVP \)]

Hi,

That's the normal behavior, see this scenario
short i = 32767 ; // the max positive value of a short aka int16
short j = 32;

the result of i + j overflow a short, so basically you have three choises:
1- ignore the overflow and do nothing, now you have an invalid result in k ,
tracking this error is a big pain!! ( I don;t really remember but I think
that this is what C/C++ does )

2- have the compiler do the check and throw an Overflow Exception, very
costly as now a lot of operations must be checked for this. C# provided a
controled way of doing this, see the checked keyword in C#

3- Automatically promote the type , this is the safer and faster way . I
think that Jave does this too ( is that right jon? )

Hope this help,

--
Ignacio Machin,
ignacio.machin AT dot.state.fl.us
Florida Department Of Transportation

"rollasoc" <ho*****@hotmail.com> wrote in message
news:uz**************@TK2MSFTNGP10.phx.gbl...

Hi,

Made me laugh so I though I'd share this.

short i = 32;
short j = 1;
short k = i + j;

gives an error
Cannot implicitly convert type 'int' to 'short'

That is a bit poo in my reckoning...

at least the following compiles.

short k = (short)(i + j);

[Jon Skeet [C# MVP]]

<"Ignacio Machin \( .NET/ C# MVP \)" <ignacio.machin AT
dot.state.fl.us>> wrote:

That's the normal behavior, see this scenario
short i = 32767 ; // the max positive value of a short aka int16
short j = 32;

the result of i + j overflow a short, so basically you have three choises:
1- ignore the overflow and do nothing, now you have an invalid result in k ,
tracking this error is a big pain!! ( I don;t really remember but I think
that this is what C/C++ does )

2- have the compiler do the check and throw an Overflow Exception, very
costly as now a lot of operations must be checked for this. C# provided a
controled way of doing this, see the checked keyword in C#

3- Automatically promote the type , this is the safer and faster way . I
think that Jave does this too ( is that right jon? )

Yup, Java does the same thing. Of course, the result of adding two ints
could overflow too - I believe it's more for efficiency reasons than to
avoid overflow, if you see what I mean. Modern processors tend to have
instructions to add 2 32-bit integers, but possibly not 16 or 8 bits...

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http://www.pobox.com/~skeet
If replying to the group, please do not mail me too