谁能告诉我这是什么大法。。。。？

//后面为输出
double test = 6 / 4;
    cout << test<<endl;   //1
    printf("%f \n", test);  //1.000000
    cout << 6 / 4 << endl; //1
    cout << (double)6 / 4 << endl; //1.5
printf("%f\n", 6 / 4); //0.000000