long x = <some value>
int y = <some value>
I want to subtract y from x , which of the following will give me different or same results
x = (int)x - y;
x = x-y
x = short(x) - short(y)
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SaiVamshi Dobbali
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3It depends on the values of x and y and if they exceed the range of short or int. Did you try it? – Jabberwocky Dec 16 '19 at 06:22
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1Before you cast, you must make sure the `long` will fit in `int` (e.g. check against `INT_MIN/INT_MAX`) and the same for `short`. (e.g. `SHRT_MIN/SHRT_MAX`) – David C. Rankin Dec 16 '19 at 06:22
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What is `short(x)` ? Did you mean `(short)x` ? Anyway - why don't you just give it a try? If you don't understand what you see then you can ask here. – Support Ukraine Dec 16 '19 at 06:37
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You could try these numbers ` x=5234000000L; y = 100; ` – Support Ukraine Dec 16 '19 at 06:41
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or read this: https://stackoverflow.com/a/28142965/4386427 – Support Ukraine Dec 16 '19 at 07:00
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1See [Implicit type promotion rules](https://stackoverflow.com/questions/46073295/implicit-type-promotion-rules). – Lundin Dec 16 '19 at 09:53
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The type long is "bigger" than int. Smaller types can be automatically casted to bigger types. So, this code:
someLongNumber + someIntNumber
is basically equal to
someLongNumber + (long)someIntNumber
So the output will be long. You don't have to cast if you want to place the result in x. However, if you want to place it in y, you must cast the x operand to int ((int)x), or the whole operation ((int)(x - y)).
CoderCharmander
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1Nitpick: the Standard guarantees that a `long` has greater _rank_ than an `int`, but both may be the same size (the only guarantee is that a `long` is at least as wide as an `int`), and on 32 bit Windows platforms `long` and `int` are in fact the same size. Also, what you are calling an "automatic cast" isn't really a cast, but is referred to in the Standard as an _implicit conversion_. – ad absurdum Dec 16 '19 at 07:33
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