【发布时间】:2014-08-01 22:30:42
【问题描述】:
这是我正在处理的程序中的几个循环。该程序似乎在printf("TEST2"); 之后停止推进。一切一目了然。我有什么遗漏吗?
我希望循环在设置开关语句中的值后重复。我知道它至少经历过一次。
#include "stdafx.h"
#include <stdlib.h>
#include <time.h>
char *names[] = { "Denise", "Inja", "Jane", "Karen", "Maggie", "Margaret", "MJ", "Queen", "Sherri", NULL }; //ptr for names, 9 nurses
const char days[5][10] = { "Monday", "Tuesday", "Wednesday", "Thursday", "Friday" };
int randomNurse();
#define total_nurses 9 //number of nurses on staff
#define days_in_week 5 //number of work days in a week
int main() {
srand(time(NULL));
int day, pos, candidate, i, j;
int slackers[4] = { 1, 1, 1, 1 }; //array that holds the selections for who isn't working
char **name_ptr = names;
/*0 = Denise, 1 = Inja, 2 = Jane, 3 = Karen, 4 = Maggie, 5 = Margaret, 6 = MJ, 7 = Queen, 8 = Sherri*/
int avail_nurses[total_nurses] = { 1, 1, 1, 1, 1, 1, 1, 1, 1 }; //holds the status of each nurse, 0 = unavailable, 1 = available
/*prints names */
int temp_counter = 1; //counter
while (*name_ptr) {
printf("%i) %s\n", temp_counter, *name_ptr);
name_ptr++;
temp_counter++;
}
/*this assumes that no more than FOUR nurses will be away on any given week*/
printf("\nEnter numbers that correspond to the nurses who won't be available for the week.\nType up to four numbers, each separated by a space.\n");
printf("When you are done, press \"Enter\".\n");
printf("If less than four nurses will be on leave, type a \"0\" in place of a selection.\n");
printf("Example: 1 2 5 0\n\n\n");
/*week selection of unavailable nurses*/
do {
printf("Who won't be here? ");
} while (scanf("%i %i %i %i", &slackers[0], &slackers[1], &slackers[2], &slackers[3]) != 4);
/*checks the selections made, and sets the available nurses to the correct value, zero if they are slacking||vacationing*/
for (int n = 0; n < 4; n++) {
int slacker = slackers[n];
if (slacker >= 1 && slacker <= 9)
avail_nurses[slacker] = -1;
}
/*-----WEEKLY_ASSIGNMENT-----*/
int pos_per_day[days_in_week] = { 5, 9, 9, 8, 5 }; //number of nurses needed each day
int selection[days_in_week][total_nurses]; //the selected nurses per day
for (i = 0; i < days_in_week; i++) {
for (j = 0; j < total_nurses; j++) {
selection[i][j] = -1; //initialize to -1 which means no nurse is selected
}
}
//fill all the days of week
for (day = 0; day < days_in_week; day++) {
for (pos = 0; pos < pos_per_day[day]; pos++) { //for every position needed that day
do {
candidate = randomNurse();
} while (!avail_nurses[candidate]); //looks for available nurses (phrasing)
avail_nurses[candidate] = 0; //change nurses status to not available
selection[day][pos] = candidate; //fill the output array with appropriate nurse
}
for (i = 0; i < total_nurses; i++) {
avail_nurses[i] = 1; //initialize the nurses status for next day use
}
for (int n = 0; n < 4; n++) { //make sure we shame the slackers...
int slacker = slackers[n];
if (slacker >= 1 && slacker <= 9)
avail_nurses[slacker] = -1;
}
}
/*-----PRINTS SCHEDULE FOR WEEK-----*/
for (i = 0; i < days_in_week; i++) {
printf("%-10s: ", days[i]);
for (j = 0; j < total_nurses; j++) {
if (selection[i][j] != -1)
printf("%-10s ", names[selection[i][j]]);
}
printf("\n");
}
return 0;
}
/*function to generate random nurse*/
int randomNurse() {
return rand() % 9; //random number 0-8, to pick nurse
}
【问题讨论】:
-
首先使用
break而不是continue。还有,AVAIL_NURSES的定义在哪里???请添加相关的代码。顺便说一句,看起来您可以简单地检查slackers[n]是否在 1 和 9 之间,然后使用AVAIL_NURSES[slackers[n]-1] = 0而不是整个switch语句。 -
唯一可能陷入无限循环的代码是
do-while块。在该块中添加printf以查看是否是问题所在。 -
该死的R萨胡!!! UI 正要这么说!
-
为什么,哦,为什么,你使用
for-switch构造?只需将整个switch语句替换为int slacker = slackers[n]; if (slacker >= 1 && slacker <= 9]) AVAIL_NURSES[slacker] = 0; -
@barakmanos 我添加了所有代码。让我知道你的想法。
标签: c loops for-loop switch-statement