la 和 addi 在 MIPS asm 中复制寄存器的区别

Question

I'm new to MIPS and confused by a concept.

I have a value 5 stored in $s5 , and I want to copy it to $a0 so that I can use li $v0, 1 to print it. I have two ways to copy.

1. addi $a0, $s5, 0
2. la $a0, 0($s5)

Either 1. or 2. can print the value 5 if I do li $v0, 1 / syscall after it ( MARS's print-integer syscall ).

But why does it work for 2.? 2. is storing the address of $s5 at $a0 , but we need a value, not an address.

Will this be automatically handled by print_integer?

Answer 1

First of all, la isn't a MIPS hardware instruction, just a pseudo-instruction implemented by the assembler, like li . (Usually with lui / addiu to construct a 32-bit symbol address in a register, if you use it the normal way as la $reg, symbol ). Look at disassembly / machine code output, eg like MARS shows you when you assemble.

You can abuse la as a move by using a register addressing mode instead of a symbol name. ( move is another pseudo-instruction; MIPS doesn't have a hardware move, you just add with $zero or an immediate 0 .)

That la could assemble to exactly that addi $a0, $s5, 0 if the assembler chose that . (Although more likely it would pick addiu; with an immediate other than 0 is has to not trap on signed overflow. In general you never want add/addi, only addu/addiu.)

2 is storing the address of $s5 at $a0, but we need a value, not an address. Will this be automatically handled by print_integer?

No, the value in $a0 is identical either way, just like if you did move $a0, $s5 like a normal person. So there's no difference in the end result for print_integer to sort out.

You can't take the address of a register. A register can hold a memory address, but you can't take the register's address.

Yes la "takes the address" of its source operand, but note that 0($s5) isn't a register, it's the syntax for a memory addressing mode that refers to the memory at the address in $s5 , like you could use with lw . The address of that memory is just $s5 .

The C equivalent is int *a0 = &*s5; where the & cancels out the unary * dereference.