In Z80 machine code, a cheap technique to initialize a buffer to a fixed value, say all blanks. So a chunk of code might look something like this.
LD HL, DESTINATION ; point to the source
LD DE, DESTINATION + 1 ; point to the destination
LD BC, DESTINATION_SIZE - 1 ; copying this many bytes
LD (HL), 0X20 ; put a seed space in the first position
LDIR ; move 1 to 2, 2 to 3...
The result being that the chunk of memory at DESTINATION is completely blank filled. I have experimented with memmove, and memcpy, and can't replicate this behavior. I expected memmove to be able to do it correctly.
Why do memmove and memcpy behave this way?
Is there any reasonable way to do this sort of array initialization?
I am already aware of char array[size] = {0} for array initialization
I am already aware that memset will do the job for single characters.
What other approaches are there to this issue?
I believe this goes to the design philosophy of C and C++. As Bjarne Stroustrup once said, one of the major guiding principles of the design of C++ is "What you don’t use, you don’t pay for". And while Dennis Ritchie may not have said it in exactly those same words, I believe that was a guiding principle informing his design of C (and the design of C by subsequent people) as well. Now you may think that if you allocate memory it should automatically be initialized to NULL's and I'd tend to agree with you. But that takes machine cycles and if you're coding in a situation where every cycle is critical, that may not be an acceptable trade-off. Basically C and C++ try to stay out of your way--hence if you want something initialized you have to do it yourself.