ori指令在此代码中的重要性是什么？

Question

# Add the first five integers 

.text 0x3000
.globl main
main:
 add   $8,$0,$0     # sum = 0
 add   $9,$0,$0     # for (i = 0; ...
loop:
 add   $8,$8,$9     # sum = sum + i;
 addi  $9,$9,1      # for (...; ...; i++
 slti  $10,$9,5     # for (...; i<5;
 bne   $10,$0,loop
end: 
 ori   $v0, $0, 10     # system call 10 for exit
 syscall               # we are out of here.

I am not quite sure what the ori instruction is doing in the end.
Once the loop has ended it has ended, so why is it significant?

Answer 1

In this case, ori simply functions as mov .
This is because ori rd, rs, imm is defined as rd = rs | imm rd = rs | imm .
When rs is the zero register, this simply means rd = imm .

If the CPU's instruction set doesn't have an actual mov instruction, a helpful assembler would translate the pseudo-instruction mov rd, imm to ori rd, $0, imm .

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As for the loop and the ending, the thing is that an assembly language has no notion of "reaching the end of the program and exiting". Suppose your last coded instruction is located at address 0xDEADBEEF. After the CPU executes the instruction, it simply moves on to the next instruction address at 0xDEADBEF0. The next instruction address merely keeps incrementing until something happens, like a trap or whatnot.

So the conventional way to terminate a program is to make a system call. This transfers execution to the OS, kills the process, and never gives execution back to the process. To make a system call, you would put in appropriate values into registers - in this case, the value 10 is a command that the system will interpret as "exit process".