Question: Consider the following assembly code: loop: movl $0, %edx movl $0, %eax jmp .L2 .L3: movq %rdi, %rcx imulq %rdx, %rcx subq %rcx, %rax addq

Consider the following assembly code:

loop: movl $0, %edx movl $0, %eax jmp .L2 .L3: movq %rdi, %rcx imulq %rdx, %rcx subq %rcx, %rax addq $3, %rdi addq $1, %rdx .L2: cmpq %rsi, %rdx jl .L3 rep ret

The preceding code was generated by compiling C code that has the following overall form.

long loop(long x, long n) { long result = ____; long i; for (i = ____; ____; ____) { result = ____; x = ____; } return result; }

You must fill in the missing parts of the C code to get a function loop that is equivalent to the generated assembly code.

Consider the following assembly code: loop: movl $0, %edx movl $0, %eax

\#include \#include long loop(long x, long n ) \{ // implement this return 0; \} int main() return 0

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