TL; DR：
printf()在按值传递时在printLotInfo()中输出垃圾。
代码
#include <stdio.h>
#include <stdlib.h>
#include <stddef.h>
#include <math.h>
#include <string.h>
typedef struct Time
{
int hour; //Hour of day
int minute; //Minute of hour
} Time;
typedef struct Car
{
char * plateNumber; //String to hold plate
char hasPermit; //True/False
Time * enteringTime; //Time Struct
int lotParkedIn; //Where is the car located
} Car;
typedef struct ParkingLot
{
int lotNumber; //Lot identifier
double hourlyRate; //$$/h
double maxCharge; //Maximum daily charge
int capacity; //How many cars can be parked in the lot?
int currentCarCount; //Tracks # of cars in lot
double revenue; //How much money has this lot made?
} ParkingLot;
// Sets the hours and minutes amount for the given time t based
// on the specified hours h. (e.g., 1.25 hours would be 1 hour
// and 15 minutes)
void setHours(Time *t, double h)
{
if(h != -1)
{
t->hour = (int) h; //Bad practice but no overflow expected here
t->minute = (h - t->hour) * 60.0; //Cast truncates h. h - t.hour is h-truncated h
}
else
{
t->hour = -1;
t->minute = -1;
}
}
// Takes two Time objects (not pointers) and computes the difference
// in time from t1 to t2 and then stores that difference in the diff
// Time (which must be a pointer)
void difference(Time t1, Time t2, Time *diff)
{
diff->hour = t2.hour - t1.hour;
diff->minute = t2.minute - t1.minute;
if(diff->minute < 0)
{
diff->hour--; //If minutes are negative, decrement hour
diff->minute += 60; //and set minutes to complement of 60.
}
}
// Initialize the car pointed to by c to have the given plate and
// hasPermit status. The car should have it’s lotParkedIn set to
// 0 and enteringTime to be -1 hours and -1 minutes.
void initializeCar(Car *c, char *plate, char hasPermit)
{
Time * t = malloc(sizeof(Time)); //Allocate memory for time object
setHours(t, -1); //Set time with call to setHours()
c->plateNumber = plate;
c->hasPermit = hasPermit; //Set variables
c->lotParkedIn = 0;
c->enteringTime = t;
}
/********************** PROBLEM SECTION ************************************/
// Initialize the lot pointed to by p to have the given number,
// capacity, hourly rate and max charge values. The currentCarCount
// and revenue should be at 0.
void initializeLot(ParkingLot * p, int num, int cap, double rate, double max)
{
p->lotNumber = num;
p->hourlyRate = rate;
p->maxCharge = max;
p->capacity = cap;
p->currentCarCount = 0;
p->revenue = 0;
printf("PRINTING IN initializeLot():\nLot is: %d Capacity is: %d Rate is: %.2lf Maximum is: %.2lf\n",
p->lotNumber, p->capacity, p->hourlyRate, p->maxCharge);
}
// Print out the parking lot parameters so that is displays as
// follows: Parking Lot #2 - rate = $3.00, capacity 6, current cars 5
void printLotInfo(ParkingLot p)
{
printf("PRINTING IN printLotInfo():\n");
printf("Parking Lot #%d - rate = $%.2lf, capacity %d, current cars %d\n",
p.lotNumber, p.capacity, p.currentCarCount);
}
/************************* END PROBLEM SECTION *****************************/
// Simulate a car entering the parking lot
// ...
void carEnters(ParkingLot * p, Car * c, int hour, int minute)
{
if((p->capacity - p->currentCarCount) <= 0)
{
printf("Car %s arrives at Lot %d at %d:%1.2d, but the lot is full\n",
c->plateNumber, p->lotNumber, hour, minute);
return;
}
double timeToSet = (hour + (minute / 60.0));
setHours(c->enteringTime, timeToSet);
c->lotParkedIn = p->lotNumber;
p->currentCarCount += 1;
printf("Car %s arrives at Lot %d at %d:%1.2d.\n", c->plateNumber,
p->lotNumber, hour, minute);
}
// Simulate a car leaving the parking lot
// ...
void carLeaves(ParkingLot * p, Car * c, int hour, int minute)
{
Time * leaveTime = malloc(sizeof(Time));
Time * timeDifference = malloc(sizeof(Time)); //Allocate memory for time object
double timeToSet = (hour + (minute / 60.0));
setHours(leaveTime, timeToSet); //Set time with call to setHours()
difference(*(c->enteringTime), *leaveTime, timeDifference);
if(c->hasPermit == 0)
{
double carRevenue = p->hourlyRate * timeDifference->hour;
if(timeDifference->minute != 0)
carRevenue += p->hourlyRate;
if(carRevenue > p->maxCharge)
carRevenue = p->maxCharge;
p->revenue += carRevenue;
printf("Car %s leaves Lot %d at %d:%1.2d paid $%.2lf.\n",
c->plateNumber, p->lotNumber, hour, minute, carRevenue);
}
else
printf("Car %s leaves Lot %d at %d:%1.2d.\n",
c->plateNumber, p->lotNumber, hour, minute);
p->currentCarCount--;
free(c->enteringTime);
free(c);
free(leaveTime);
free(timeDifference);
}
/*BEGIN PROFESSOR'S CODE*/
int main() {
Car car1, car2, car3, car4, car5, car6, car7, car8, car9;
ParkingLot p1, p2;
// Set up 9 cars
initializeCar(&car1, "ABC 123", 0);
initializeCar(&car2, "ABC 124", 0);
initializeCar(&car3, "ABD 314", 0);
initializeCar(&car4, "ADE 901", 0);
initializeCar(&car5, "AFR 304", 0);
initializeCar(&car6, "AGD 888", 0);
initializeCar(&car7, "AAA 111", 0);
initializeCar(&car8, "ABB 001", 0);
initializeCar(&car9, "XYZ 678", 1);
// Set up two parking lots
initializeLot(&p1, 1, 4, 5.5, 20.0);
initializeLot(&p2, 2, 6, 3.0, 12.0);
printLotInfo(p1);
printLotInfo(p2);
printf("\n");
// Simulate cars entering the lots
carEnters(&p1, &car1, 7, 15);
carEnters(&p1, &car2, 7, 25);
carEnters(&p2, &car3, 8, 0);
carEnters(&p2, &car4, 8, 10);
carEnters(&p1, &car5, 8, 15);
carEnters(&p1, &car6, 8, 20);
carEnters(&p1, &car7, 8, 30);
carEnters(&p2, &car7, 8, 32);
carEnters(&p2, &car8, 8, 50);
carEnters(&p2, &car9, 8, 55);
printf("\n");
printLotInfo(p1);
printLotInfo(p2);
printf("\n");
// Simulate cars leaving the lots
carLeaves(&p2, &car4, 9, 0);
carLeaves(&p1, &car2, 9, 5);
carLeaves(&p1, &car6, 10, 0);
carLeaves(&p1, &car1, 10, 30);
carLeaves(&p2, &car8, 13, 0);
carLeaves(&p2, &car9, 15, 15);
carEnters(&p1, &car8, 17, 10);
carLeaves(&p1, &car5, 17, 50);
carLeaves(&p2, &car7, 18, 0);
carLeaves(&p2, &car3, 18, 15);
carLeaves(&p1, &car8, 20, 55);
printf("\n");
printLotInfo(p1);
printLotInfo(p2);
printf("\n");
// Display the total revenue
printf("Total revenue of Lot 1 is $%4.2f\n", p1.revenue);
printf("Total revenue of Lot 2 is $%4.2f\n", p2.revenue);
}
以上代码用于分配。我们获得了功能原型，并负责使它们执行与停车场结构有关的某些动作。数据操作实际上按预期方式工作，在程序结束时为两个批次提供了预期的输出。
这使我认为printf()引起了麻烦，而不是潜在的价值。专用printLotInfo()函数和printf中的临时initializeLot()语句之间的唯一区别是，一个从struct *打印，而另一个从结构打印。
我花了几个小时研究这个问题，并且遇到了很多线程，展示了如何从struct *中的函数或main结构中进行打印，但是我没有找到一个很好的例子在单独的函数调用中从结构打印。我认为我缺少明显的东西，例如不正确的解除引用运算符或类似的东西。
答案 0 :(得分：5)
您缺少一个参数：
printf("Parking Lot #%d - rate = $%.2lf, capacity %d, current cars %d\n",
p.lotNumber, p.capacity, p.currentCarCount);
您的格式字符串具有4个格式说明符，但您仅提供3个参数。结果，第二个格式说明符%f在其中传递double的情况下寻找int，而第四个格式说明者在 no的情况下寻找int 参数已传递。使用错误的格式说明符或未提供足够的参数会调用undefined behavior。
根据需要在参数列表的hourlyRate字段中输入
printf("Parking Lot #%d - rate = $%.2lf, capacity %d, current cars %d\n",
p.lotNumber, p.hourlyRate, p.capacity, p.currentCarCount);
答案 1 :(得分：-1)
与该问题不直接相关，但是经过一些调整，该程序可以变得更加健壮，并且也是有效的C ++程序。这有助于发现一些改进C程序的东西。
使Car保持指向const字符的指针。
typedef struct Car
{
char const *plateNumber; //String to hold plate
...
对initializeCar的界面进行了相同的更改：
void initializeCar(Car *c, char const *plate, char hasPermit)
{
Time * t =(Time*) malloc(sizeof(Time)); //Allocate memory for time object
将malloc的返回值设置为所需的类型。 （请参见上方和下方。）
void carLeaves(ParkingLot * p, Car * c, int hour, int minute)
{
Time * leaveTime = (Time*)malloc(sizeof(Time));
Time * timeDifference = (Time*)malloc(sizeof(Time)); //Allocate memory for time object
最后，让main的签名故意说它需要零个参数。
int main(void) {
在没有void的旧接口的情况下，它需要接受零个到无限个参数。 （在C ++ main()中已经意味着零个参数。）