Good! I started to study at little assembly for MIPS. Can anyone explain to me how this line of code works, ld r4, A (r0), where r0 has the same meaning as $ zero?
.data
A: .word 10
B: .word 8
C: .word 0
.text
main:
ld r4,A(r0)
ld r5,B(r0)
dadd r3, r4, r5
sd r3,C(r0)
halt
By the type of statements used ( ld and dadd ), I believe this is MIPS-64 ( font that says dadd is present in MIPS-64).
I have no experience with MIPS-64 and have worked little with the MIPS-32 Assembly, but this is my reading of the code:
ld is load (64-bit) and dadd is signed add (64-bit).
This code begins by declaring three variables: A, B, and C with the initial values 10 , 8 , and 0 . These variables are declared the size of a word. I'm not sure if this is 32 or 64 bits, but the code that follows only makes sense if they are 64-bit.
Log reads $zero ( r0 ) give the value 0 and written in this log are ignored.
ld X, Y(Z) loads in register X the value in position Y + Z (this works when accessing constant indexes of arrays). Since Z is zero, load the position directly.
What this does is:
r4 . r5 . r3 . r3 in global variable C. That is, C = A + B . Therefore, C ends with the value 18 .
The equivalent C code is:
#include <stdint.h>
int64_t A, B, C;
int main()
{
C = A + B;
// Ou o seguinte, que é equivalente:
// (&C)[0] = (&A)[0] + (&B)[0];
halt(); // Talvez implementado como for (;;) {}
// Embora imagino que o halt seja mais eficiente
}