College

Translate the following LEGv8 assembly into C, assuming that:
- X19 = f
- X20 = g
- X21 = h
- X22 = &a

```
SUB X9, X20, X21
CBZ X9, ELSE
LSL X9, X20, #3
ADD X9, X22, X9
LDUR X19, [X9, #0]
B END
ELSE: LDUR X19, [X22, #0]
END: SUB X21, X19, X20
```

The translated C code is:

```c
if (g != h) {
f = *(a + (g << 3));
} else {
f = *a;
}
h = f - g;
```

Answer :

Final answer:

The LEGv8 assembly code provided is translated into a C snippet that uses if-else to determine whether 'g' equals 'h', and then computes 'f' based on the result, with 'h' being updated to 'f - g'.

Explanation:

The LEGv8 assembly code given can be translated into a C program as follows:

int f, g, h, a, *pa;
pa = &a;
if (g == h) {
f = *pa;
} else {
int temp = g;
temp <<= 3;
f = *(pa + temp);
}
h = f - g;

The operations in the assembly code are translated into C operations where X19, X20, X21, and X22 correspond to variables f, g, h, and the address of a, respectively. The SUB (subtract) instruction calculates the difference between g and h, then the CBZ (compare and branch if zero) instruction checks if the result is zero (meaning g is equal to h) and branches accordingly. The LSL (logical shift left) and ADD instructions calculate an offset used as part of an address for a memory access with LDUR, and finally, h gets the result of f minus g.

Learn more about LEGv8 Assembly Translation here:

https://brainly.com/question/36745644

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