C ++至MIPS汇编

Question

if i transfer this code to mips.

int arr[3];

cin>>arr[0];

cin>>arr[1];

arr[1]+=arr[0];

cin>>arr[2];

arr[2]+=arr[1];

if i have value of arr[0] in $t0 ,arr[1] in $t1 and address of the arr in $s0.

in this line

arr[1]+=arr[0];

what i should do from this ? use $t1 and $t0 direct like this

add $t1,$t1,$t0

or i should get the value again from the memory in a registers and do the add instruction like this:

lw $s1,($S0)

lw $s2,4($S0)

add $s2,$s2,$s1

what the compiler do ?

Answer 1

There is no need to get the values from the addresses again if you have properly loaded the values into $t0 and $t1 .

Compilers are very complex and what a compiler produces depends on many things, like optimization as Tilo mentioned. The best way to see is to compile it yourself and look at the code produced.