目标：斐波那契数列 前30位 循环100万次

位数多了，大数的话，JS还是C代码写起来很麻烦，嗯，更别说汇编了

代码：

function jstest(){
    var time = new Date();
    var a,b,res;
    for(var t =0;t<1000000;t++) {
        a = 0;
        b = 1;
        for (var i = 0; i < 30; i++) {
            res = a + b;
            a = b;
            b = res;
        }
    }
    console.log("value is:",res);
    time = new Date().getTime() - time.getTime();
    console.log('js',time+'ms');
}
function addon(){
    var time = new Date();
    x.fib();
    time = new Date().getTime() - time.getTime();
    console.log('addon c++',time+'ms');
}
function assb(){
    var time = new Date();
    x.assb();
    time = new Date().getTime() - time.getTime();
    console.log('addon c++ / inline asm',time+'ms');
}
jstest();
addon();
assb();

c++

    Isolate* isolate = Isolate::GetCurrent();
    HandleScope scope(isolate);
    long a,b,c;
    for(int i=0;i<1000000;i++ ){
	a = 0;
        b = 1;
        for(int t=0;t<30;t++){
            c=a+b;
            a = b;
            b = c;
        }
    }
    std::cout<<"value is: "<<c<<std::endl;

c++ inline asm

    Isolate* isolate = Isolate::GetCurrent();
    HandleScope scope(isolate);
    int b;
    __asm__ __volatile__(
        "movl	$100000, %%edi\n\t"
        "jmp	.L2\n\t"
        ".L3:\n\t"
        	"movq	%%rdx, %%rcx\n\t"
        	"movq	%%rsi, %%rdx\n\t"
        ".L5:\n\t"
            "leaq	(%%rcx,%%rdx), %%rsi\n\t"
        	"subl	$1, %%eax\n\t"
            "jne	.L3\n\t"
            "subl	$1, %%edi\n\t"
            "je	.L7\n\t"
        ".L2:\n\t"
        	"movl	$30, %%eax\n\t"
            "movl	$1, %%edx\n\t"
            "movl	$1, %%ecx\n\t"
        	"jmp	.L5\n\t"
        ".L7:\n\t"
        	"movq	%%rdx,%%rax\n\t"
        	"movl   %%eax,%0\n\t"
        :"=a" (b)
    );
    std::cout<<"value is: "<<b<<std::endl;

结果

value is: 1346269 js 115ms value is: 1346269 addon c++ 31ms value is: 1346269 addon c++ / inline asm 3ms

配置

Intel® Xeon® CPU E5-2430 0 @ 2.20GHz -单核 512MB 阿里云

后话

我开始自己写了LOOP体汇编。居然跟JS的性能差不多！后来直接反编译C。得到了上面的性能代码，瀑布汗。 可挑战CPU密集型业务。