管道传输时,C++程序的性能更佳

Question

十年来我没有做过任何编程.我想回到它,所以我把这个毫无意义的程序作为练习.描述它的作用的最简单方法是输出我的--help codeblock:

./prng_bench --help

./prng_bench: usage: ./prng_bench $N $B [$T]

   This program will generate an N digit base(B) random number until
all N digits are the same. 

Once a repeating N digit base(B) number is found, the following statistics are displayed:
  -Decimal value of all N digits.
  -Time & number of tries taken to randomly find.

Optionally, this process is repeated T times. 
   When running multiple repititions, averages for all N digit base(B)
numbers are displayed at the end, as well as total time and total tries.

我的"问题"是,当问题"容易"时,比如一个3位数的基数10,并且我做了大量的传递,"总时间"在用管道传输到grep时较少.即:

命令 命令| grep:

./prng_bench 3 10 999999 ; ./prng_bench 3 10 999999|grep took

....
Pass# 999999: All 3 base(10) digits =  3 base(10).   Time:    0.00005 secs.   Tries: 23
It took 191.86701 secs & 99947208 tries to find 999999 repeating 3 digit base(10) numbers.
An average of 0.00019 secs & 99 tries was needed to find each one. 

It took 159.32355 secs & 99947208 tries to find 999999 repeating 3 digit base(10) numbers.

如果我多次运行相同的命令没有grep时间总是非常接近.我现在正在使用srand(1234)进行测试.我对clock_gettime()的启动和停止调用之间的代码不涉及任何流操作,这显然会影响时间.我意识到这是一种无用的练习,但我想知道它为什么会这样.以下是该计划的核心.如果有人想编译和测试,这里是DB的完整源代码的链接.https://www.dropbox.com/s/bczggar2pqzp9g1/prng_bench.cpp clock_gettime()需要-lrt.

for (int pass_num=1; pass_num<=passes; pass_num++) {   //Executes $passes # of times.
  clock_gettime(CLOCK_PROCESS_CPUTIME_ID, &temp_time);  //get time
  start_time = timetodouble(temp_time);                //convert time to double, store as start_time
  for(i=1, tries=0; i!=0; tries++) {    //loops until 'comparison for' fully completes. counts reps as 'tries'.  <------------
    for (i=0; i<Ndigits; i++)      //Move forward through array.                                                              |
      results[i]=(rand()%base);    //assign random num of base to element (digit).                                            |
    /*for (i=0; i<Ndigits; i++)     //---Debug Lines---------------                                                           |
      std::cout<<" "<<results[i];   //---a LOT of output.----------                                                           |
    std::cout << "\n";              //---Comment/decoment to disable/enable.*/   //                                           |
    for (i=Ndigits-1; i>0 && results[i]==results[0]; i--); //Move through array, != element breaks & i!=0, new digits drawn. -|
  }                                                        //If all are equal i will be 0, nested for condition satisfied.  -|
  clock_gettime(CLOCK_PROCESS_CPUTIME_ID, &temp_time);  //get time
  draw_time = (timetodouble(temp_time) - start_time);  //convert time to dbl, subtract start_time, set draw_time to diff.
  total_time += draw_time;    //add time for this pass to total.
  total_tries += tries;       //add tries for this pass to total.
  /*Formated output for each pass:
    Pass# ---: All -- base(--) digits = -- base(10)   Time:   ----.---- secs.    Tries: ----- (LINE) */
  std::cout<<"Pass# "<<std::setw(width_pass)<<pass_num<<": All "<<Ndigits<<" base("<<base<<") digits = "
           <<std::setw(width_base)<<results[0]<<" base(10).   Time: "<<std::setw(width_time)<<draw_time
           <<" secs.   Tries: "<<tries<<"\n";
}
if(passes==1) return 0;        //No need for totals and averages of 1 pass.
/* It took ----.---- secs & ------ tries to find --- repeating -- digit base(--) numbers. (LINE)
 An average of ---.---- secs & ---- tries was needed to find each one. (LINE)(LINE) */
 std::cout<<"It took "<<total_time<<" secs & "<<total_tries<<" tries to find "
          <<passes<<" repeating "<<Ndigits<<" digit base("<<base<<") numbers.\n"
          <<"An average of "<<total_time/passes<<" secs & "<<total_tries/passes
          <<" tries was needed to find each one. \n\n";
return 0;

Answer 1

与管道相比或在没有打印的情况下运行时,打印到屏幕上的速度非常慢.管道到grep会让你无法做到这一点.