测试 sum = sum + i * i 循环1亿次
为什么不是单纯的加法呢，因为发现当i到一定程度的时候，c++的计算时间都是0，可能使用了 优化 sum =（i+1)*i/2
等有空分析c++汇编代码
减少c++的另类优化，所以用的加 i * i，公平起见
1. lua代码
#!/usr/bin/lua
require("os")
local starttime = os.clock()
local length = 100000000
local sum = 0
for i = 1, length do
sum = sum + i * i
end
print("sum = ", sum)
local endtime = os.clock()
print(string.format("cost time : %.4f", endtime - starttime))
cost time : 1.4200
cost time : 1.4400
cost time : 1.4200
cost time : 1.4600
cost time : 1.4300
平均时间 7.17/6 = 1.434
2.cpp
test.cpp
#include <stdio.h>
#include <time.h>
void test_add(unsigned int length)
{
unsigned long long sum = 0;
for(long i=1;i<length; ++i)
sum += i*i;
printf("sum = %lu\n", sum);
}
class CalcTime
{
public:
CalcTime()
{
m_start = clock();
}
~CalcTime()
{
unsigned long long end = clock();
printf("cost time = %f\n", ((double)(end - m_start))/CLOCKS_PER_SEC);
}
private:
unsigned long long m_start = 0;
};
int main()
{
printf("CLOCKS_PER_SEC = %d\n", CLOCKS_PER_SEC);
{
CalcTime calc;
test_add(100000000);
}
return 0;
}
Makefile
test: test.cpp
g++ -g -O2 --std=c++11 test.cpp -o test
clean:
rm test
cost time = 0.050000
cost time = 0.050000
cost time = 0.050000
cost time = 0.060000
cost time = 0.050000
平均时间 0.052
3. Go
test_go.go
package main
import "time"
import "fmt"
func test_add(len uint) {
var sum uint64 = 0
var i uint = 1
for ; i < len; i++ {
sum += (uint64)(i * i)
}
fmt.Printf("sum = %v\n", sum)
}
func main() {
start := time.Now()
test_add(100000000)
end := time.Now()
dif := end.Sub(start)
fmt.Printf("cost time: %v\n", dif)
}
cost time: 56.804357ms
cost time: 57.604325ms
cost time: 57.030147ms
cost time: 55.54873ms
cost time: 58.058417ms
平均时间=56.98ms
结论：
C++ / Go /Lua 效率对比
52 ：56.98 ：1434
1：1.0958：27.5769
网友评论