斐波那契数列是一个满足递推关系的数列,如:1 1 2 3 5 8 ...
其前两项为1,第3项开始,每一项都是其前两项之和。
用Go实现一个简单的斐波那契计算逻辑
func fib(n int) int {
if n == 1 || n == 2 {
return 1
}
return fib(n-1) + fib(n-2)
}
我们将其改进一下,用更简单的方式描述,同时把变量的定义提到前边,并将返回的逻辑拿到函数末尾。
func fib1(n int) int {
var tmp, res, res1, res2 int
if n == 1 {
goto return1
}
if n == 2 {
goto return1
}
tmp = n - 1
res1 = fib1(tmp)
tmp = n - 2
res2 = fib1(tmp)
res = res1 + res2
return res
return1:
return 1
}
继续改进
func fib2(n int) int {
var tmp, res, res1, res2 int
if n == 1 {
goto return1
}
if n == 2 {
goto return1
}
tmp = n
tmp -= 1
res1 = fib2(tmp)
tmp = n
tmp -= 2
res2 = fib2(tmp)
res = res1
res += res2
return res
return1:
return 1
}
继续改进,复用变量
func fib3(n int) int {
var res, res1 int
if n == 1 {
goto return1
}
if n == 2 {
goto return1
}
res1 = n
res1 -= 1
res = fib3(res1)
res1 = n
res1 -= 2
res1 = fib3(res1)
res += res1
return res
return1:
return 1
}
继续改进
func fib4(r0 int) int {
var r1, r2 int
if r0 <= 2 {
goto return1
}
r2 = r0
r2 -= 1
r1 = fib4(r2)
r2 = r0
r2 -= 2
r2 = fib4(r2)
r1 += r2
return r1
return1:
return 1
}
汇编函数接口定义
func fib5(n int) int
汇编函数实现文章来源:https://www.toymoban.com/news/detail-807170.html
TEXT ·fib5(SB),$40-16
MOVD R0, tmp-24(SP);MOVD R1, tmp-16(SP);MOVD R2, tmp-8(SP) // save context
MOVD n+0(FP), R0; MOVD $0, R1; MOVD $0, R2 // R0: load value from stack memory
CMP $2, R0; BLE LABEL_RETURN1 // if R0 <= 2 {goto LABEL_RETURN1}
MOVD R0, R1; SUB $1, R1 // R1 = R0 - 1
MOVD R1, tmp-40(SP); CALL ·fib5(SB); MOVD tmp-32(SP), R1 // R1 = fib5(R1)
MOVD R0, R2; SUB $2, R2 // R2 = R0 - 2
MOVD R2, tmp-40(SP); CALL ·fib5(SB); MOVD tmp-32(SP), R2 // R2 = fib5(R2)
ADD R1, R2; MOVD R2, R0 // R0 = R1 + R2
JMP LABEL_END
LABEL_RETURN1:
MOVD $1, R0 // R0 = 1
LABEL_END:
MOVD R0, n+8(FP) // set return value: R0
MOVD tmp-24(SP), R0; MOVD tmp-16(SP), R1; MOVD tmp-8(SP), R2 // recovery context
RET
结果:
文章来源地址https://www.toymoban.com/news/detail-807170.html
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