byte + byte = int: why?

[Ali Çehreli]

On 1/18/19 9:09 AM, Mek101 wrote:

 >      source/hash.d(11,25): Error: cannot implicitly convert expression
 > `cast(int)temp[fowardI] + cast(int)temp[backwardI]` of type `int` to 
`byte`

Others suggested casting as a solution which works but it will hide 
potential errors.

Depending on the situation, you may want to use std.conv.to, which does 
a value range check and throws an exception to prevent an error:

byte foo(byte a, byte b) {
   import std.conv : to;
   return (a + b).to!byte;
}

void main() {
   foo(42, 42);    // Works
   foo(100, 100);  // Throws ConvOverflowException
}

Of course, because of the checks, 'to' is slower than casting blindly.

Ali