短程序在 C 中挂起

Question

每次在 UNIX 中运行此代码时，它都会挂起并显示新行而不提示。它编译没有错误。有人知道这里发生了什么吗？

/*
This program uses a Monte Carlo simulation to maximize widget profits.
 */

#include <stdio.h>
#include <stdlib.h>

// constant values used in the simulation
#define MIN_WIDGETS 1000 // minimum widgets per month
#define MAX_WIDGETS 1500 // maximum widgets per month

#define MIN_COST 0.25 // minimum cost per widget
#define MAX_COST 0.82 // maximum cost per widget

#define MIN_CONVERSION 0.01 // minimum converstion rate
#define MAX_CONVERSION 0.05 // maximum converstion rate

#define MIN_PROFIT 38 // minimum profit per sale
#define MAX_PROFIT 43 // maximum profit per sale

#define FIXED_OVERHEAD 400 // fixed overhead cost

#define NUM_SIMULATIONS 100 // number of simulations performed

void main()
{

  // to keep track of inputs in highest-profit simulation
  int bestW = 0; // widgets per month
  float bestC = 0; // cost per widget
  float bestR = 0; // rate of converstion
  float bestP = 0; // profit per sale
  float bestProfit = 0; // highest profit

  srand(time(0)); // initialize the random number generator
  float h = FIXED_OVERHEAD;

  int i; // loop index 
  for(i = 0; i < NUM_SIMULATIONS; i++)
    {
        // initialize inputs for this individual simulation
        int w = 0; // quantity of widgets bought
        float c = 0; // cost of a widget
        float r = 0; // conversion rate
        float p = 0; // profit per sale
        float profit = 0; // profit

        // simulate quantity of widgets bought per month, between MIN_WIDGETS and MAX_WIDGETS
        w = random() % (MAX_WIDGETS + 1); // to set the maximum value of w at MAX_WIDGETS
        while (w < MIN_WIDGETS) {
            w = random() % (MAX_WIDGETS + 1); // fetch new random number that may fit parameters
        }

        // simulate cost per widget, between MIN_COST and MAX_COST
        c = random() % 100; // to convert random number into an integer between 0 and 99
        while (c < (MIN_COST*100) || c > (MAX_COST*100)) {
            c = random() % 100; // fetch new random number that may fit parameters
        }
        c = c / 100.0; // convert cost back from cents into dollars

        // simulate conversion rate, between MIN_CONVERSION and MAX_CONVERSION
        r = random() % 100; // to convert random number into an integer between 0 and 99
        while (r < (MIN_CONVERSION*100) || r > (MAX_CONVERSION*100)) {
            r = random() % 100; // fetch new random number that may fit parameters
        }
        r = r / 10.0; // convert back into fraction

        // simulate profit per sale, between MIN_PROFIT and MAX_PROFIT
        p = random() % ((MAX_PROFIT + 1)*100); // to convert random number into an integer between 0 and 4300
        while (p < MIN_PROFIT*100) {
            p = random() % (MAX_PROFIT + 1); // fetch new random number that may fit parameters
        }
        p = p / 100.0; // convert back into floating point with two decimal places after period

        profit = (w * r * p) - (h + c * w);

        printf("Current profit is $%.2f, with %d widgets at a %.2f cost per widget with a %.1f conversion rate and %.2f profit/sale.\n",
            profit, w, c, r, p);

        if (profit > bestProfit) {
            bestW = w;
            bestC = c;
            bestR = r;
            bestP = p;
            bestProfit = profit;
        }

    }

  printf("Maximum profit is $%.2f, with %d widgets at a %.2f cost per widget with a %.1f conversion rate and %.2f profit/sale.\n",
    bestProfit, bestW, bestC, bestR, bestP);

}

Answer 1

我认为错误就在这里：

while (p < MIN_PROFIT*100) {
    p = random() % (MAX_PROFIT + 1); // fetch new random number that may fit parameters
}

如果进入循环体，p 将被设置为 0 到 MAX_PROFIT (= 43) 之间的数字。它永远不会大于或等于 MIN_PROFIT*100 (= 3800)，因此它将进入无限循环。

作为旁注，您可能需要考虑在此处使用 do { } while 循环，而不是使用 while 循环并编写循环体两次。编写两次代码是一种极容易出错的方法，因为当您更改其中一个实现时，您始终必须记住更改另一个实现。如果您忘记更新两者，则会引入错误。这似乎就是这里发生的事情。

Answer 2

很难从那堵代码墙上看出，但它要么花费的时间比您想象的要长，要么您的许多 while 语句之一不正确。

您可以替换 while 语句，并通过获取 0 和各种最小和最大参数之间的差异之间的随机数来加快整个过程，然后将其添加到最小参数，而不是您目前正在做的事情最大参数值的随机数，如果低于最小值则重试。例如，对于您的小部件，您将丢弃 2/3 的结果。