C - For 循环仅针对我修改的数组运行一次

Question

首先，对可能的简单问题和糟糕的代码表示歉意，就像大多数发布问题的人一样，我对相关语言非常陌生。

代码用途:

创建一个模拟进程调度的程序。这个特定实例是模仿先来先服务调度。所讨论的数据集都是有序的(例如 PID 的 0 到 5 都按顺序到达)和无序的(例如 3 可以在 1 之前到达等)。

为了实现这一目标，我尝试在到达时间数组中找到最小的条目，以确保条目按顺序处理。因此，我在数组运行 for 循环时修改数组，更改最小值，这样它就不会运行两次(我相信这可能是我的问题，但我不确定)

我遇到问题的特定循环:

for (i = 1; i < processes; i++) {
    atperm[position] = at[position];
    smallestentry();
    wt[position] = btt - at[position]; //wait time = total burst time - arrival time
    btt += bt[position]; //total burst time = total burst time + burst time
    awt += wt[position]; //average wait time = average wait time + wait time
    tat[position] = wt[position] + bt[position]; //turn around time = wait time + burst time
    atat += tat[position]; //average turn around time = average turn around time + turn around time
    changeline(position);
}

完整代码:

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

/* 
 * WT - Wait Time (TAT -BT)
 * BT - Bust Time
 * AT - Arrival Time
 * TAT - Turn Around Time (CT - AT)
 * AWT - Average Wait Time
 * ATAT - Average Turn Around Time
 */

int wt[10], bt[10], at[10], atperm[10], tat[10], processes, smallest, i, position;
float awt, atat;

void input() {
    printf("Enter number of processes:\n");
    scanf("%d", &processes);
    int i;
    for (i = 0; i < processes; i++) {
        printf("Enter burst time of process %d:", i + 1); //increase i by 1 but dont save
        scanf("%d", &bt[i]);
        printf("Enter arrival time of process %d:", i + 1);
        scanf("%d", &at[i]);
    }
}

void changeline(int position) {
    at[position] = 999;

    printf("AT\tBT\tWT\tTAT\tAT\n");
    for (i = 0; i < processes; i++) {
        printf("%3d\t%3d\t%3d\t%3d\t%3d\n", at[i], bt[i], wt[i], tat[i], atperm[i]);
    }
}

void smallestentry() {
    smallest = at[0];
    for (i = 0; i < processes; i++) {
        if (smallest > at[i]) {
            smallest = at[i];
            position = i;
        }
    }
}

void calculate() {
    smallestentry();
    wt[position] = 0; //sets the first entry in wait time array to equal 0
    atat = tat[position] = bt[position]; //sets the first entries in atat and tat array to equal first bt entry
    int btt = bt[position]; //to store total burst time sum
    atperm[position] = at[position];
    changeline(position);

    for (i = 1; i < processes; i++) {
        atperm[position] = at[position];
        smallestentry();
        wt[position] = btt - at[position]; //wait time = total burst time - arrival time
        btt += bt[position]; //total burst time = total burst time + burst time
        awt += wt[position]; //average wait time = average wait time + wait time
        tat[position] = wt[position] + bt[position]; //turn around time = wait time + burst time
        atat += tat[position]; //average turn around time = average turn around time + turn around time
        changeline(position);
    }
    atat /= processes; // atat = atat / processes
    awt /= processes; // awt = awt / processses
}

int main() {
    printf("FCFS CPU Scheduling Algorithm\n");
    input();
    calculate();
}

Answer 1

你唯一的问题是你使用 i 作为全局变量。

int wt[10], bt[10], at[10], atperm[10], tat[10], processes, smallest, i, position;

在循环中间你调用了函数smallestentry()，它也使用了i:

void smallestentry() {
smallest = at[0];
    for (i = 0; i < processes; i++) {
        if (smallest > at[i]) {
            smallest = at[i];
            position = i;
        }
    }
}

此函数结束后，i 的值发生更改，因此循环仅运行一次。

我通过将 printf("%d",i); 放入循环中发现了问题。

您可以通过将 i 更改为函数内的局部变量来解决此问题。

(抱歉有任何错误的措辞，这是我的第一个答案。)