Physics for Science and Engineering II

PHYS | 1302 | Physics for Science and Engineering II | homework | 

Calculus 2

MATH | 1372 | Calculus 2 | homework | 

Machine Architecture

A. Diagram how this instruction sequence would be decoded into operations, and show how the data dependencies between them would create a critical path of operations, in the style of Figure 5.13 and Figure 5.14 as shown below. %rax %rdx %xmm0 %rax %rdx %xmm0 load mul add cmp jne %rax %rdx load mul add cmp jne %xmm0 %rdx %xmm0 B. For data type double , what lower bound on the CPE is determined by the critical path? C. Assuming similar instruction sequences for the integer code as well, what lower bound on the CPE is determined by the critical path for integer data? D. Explain how the ﬂoating-point versions can have CPEs of 3.00, even though the multiplication oper- ation requires 5 clock cycles. Problem 4: (related to textbook problem 5.14) A. Write a version of the inner product procedure described in Problem 3, with the following optimiza- tions: six-way loop unrolling, six parallel accumulators and re-association. Write one version for integer data ( long ), and one for ﬂoating point ( double ), both for x86-64. B. For each data type, list the instructions on each critical path with and without re-association (four paths in all). 5 Problem 5: The ﬁle f.c is on the left below, and it compiles to an object ﬁle f.o shown disassembled on the right: int x; int * f(void) { return &x; } 0000000000000000 <f>: 0: 55 push %rbp 1: 48 89 e5 mov %rsp,%rbp 4: b8 00 00 00 00 mov $0x0,%eax 5: R_X86_64_32 x 9: 5d pop %rbp a: c3 retq The ﬁle g.c is on the left below, and it compiles to an object ﬁle g.o shown disassembled on the right: static int y; int * g(void) { return &y; } 0000000000000000 <g>: 0: 55 push %rbp 1: 48 89 e5 mov %rsp,%rbp 4: b8 00 00 00 00 mov $0x0,%eax 5: R_X86_64_32 .bss 9: 5d pop %rbp a: c3 retq The ﬁle main.c looks like this (except we’ve left some blanks, each ﬁlled in with a single identiﬁer): extern int x; extern int y; int * f(void); int * g(void); int printf(const char * fmt, ...); int main(void) { printf("%d %d\n", ____ + * ____(), * ____()); } and is compiled to an object ﬁle main.o like this (we’ve left out relocation information): 0000000000000000 <main>: 0: 55 push %rbp 1: 48 89 e5 mov %rsp,%rbp 4: 53 push %rbx 5: 48 83 ec 08 sub $0x8,%rsp 9: e8 00 00 00 00 callq e <main+0xe> e: 8b 18 mov (%rax),%ebx 10: e8 00 00 00 00 callq 15 <main+0x15> 15: 8b 10 mov (%rax),%edx 17: 8b 05 00 00 00 00 mov 0x0(%rip),%eax 1d: 01 d0 add %edx,%eax 1f: 89 da mov %ebx,%edx 21: 89 c6 mov %eax,%esi 23: bf 00 00 00 00 mov $0x0,%edi 28: b8 00 00 00 00 mov $0x0,%eax 2d: e8 00 00 00 00 callq 32 <main+0x32> 32: b8 00 00 00 00 mov $0x0,%eax 37: 48 83 c4 08 add $0x8,%rsp 3b: 5b pop %rbx 3c: 5d pop %rbp 3d: c3 retq 6

CSCI | 2021 | Machine Architecture | lecture notes | A. Diagram how this instructio

hw4-soln.pdf-CSci 2021, Spring 2016 Homework Assignment

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