How does lw know how to load word sizes?

[Ryan Tio]

# Lecture Question

Week : 3

Slide deck name: Control-Flow

Question: For the array example, I was wondering how we dealt with arrays that maybe contained a complex type or a struct, how do we know that loading a single word size into the register will be enough to contain the whole object, or will it always be a pointer to the object?

[Ryan Tio]

I mean something like 

struct names {

...

};

struct names arr[10]; 

and we wanted to load say arr[5] into the registry using something like 

lw x12, 0(x14)

How does lw know that the object will fit into the single word size?

[Arrvindh Shriraman]

Short answer: you can't. A single C statement will get lowered to multiple assembly instructions by the compiler

Two reasons

- Registers are only 1 word wide (see Registers video)

- You do not have access to those types of load instructions. 

You can only operate on one word at a time.

See here for instrance. 

https://godbolt.org/z/EzGY88

Check out the copy function. I am saying record[0] = record[1], which means internally I need multiple lw and sw to copy from one location in memory to another one word at a time

Modify the code and see how that leads to different instruction sequences. 

(Small correction: It is not registry. It is either registers or register file).

-----------------------------------

If you have not tried godbolt before then my videos should include a short description.

https://www.youtube.com/watch?v=VkqSWIy9DjI&feature=youtu.be 

This video also includes more details. (https://www.youtube.com/watch?v=4_HL3PH4wDg)

- Make sure you select the RISC-V compiler shown in my link if you reload the page.

  if you select a different compiler then the assembly for a different processor will be generated

- Make sure you set the command line arg (like I have done) to -O1. This makes the code more readable.