why i can't make operators on 8- and 16-Bit Integrals

Question

short x = 1, y = 1;
short z = x + y; // Compile-time error

So what's the error? – Maria Ines Parnisari Apr 14 '16 at 20:08 — Maria Ines Parnisari, Apr 14 '16 at 20:08

score 1 · Answer 1 · answered Apr 14 '16 at 20:09

1

The explanation is given here.

The following assignment statement will produce a compilation error, because the arithmetic expression on the right-hand side of the assignment operator evaluates to int by default.

Since there is no implicit conversion from int to short, you have to do

short z = (short) (x + y);

answered Apr 14 '16 at 20:09

Paul Boddington

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thank you, so why we sometimes use short and byte variables ? – Nour Ahmed Apr 14 '16 at 20:15
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byte arrays are very useful for IO. A byte represents a small packet of data. As far as I can tell there is no point in `short` in C# or java at all. I never use it. – Paul Boddington Apr 14 '16 at 20:17
@NourAhmed This is probably a better answer. http://stackoverflow.com/a/1541484/3973077 – Paul Boddington Apr 14 '16 at 20:20
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It seems the road is still long to learn the language I love to understand, thanks to you my friend has informed me – Nour Ahmed Apr 14 '16 at 20:22
No problem. I'm glad it helped. – Paul Boddington Apr 14 '16 at 20:22

score 0 · Answer 2 · answered Apr 14 '16 at 20:11

You could if you cast the result, the point is: shorts are 16 bits number, hence a short can have a value between -32.768 and 32.767, but when you do short x = 5; you are actually assigning an integer as literal

Adding 2 integers can for sure overflow the shorts capacity, hence the complains of the ide...

why i can't make operators on 8- and 16-Bit Integrals

2 Answers2