c - 值同步无法正常工作

Question

我的程序的此链接中有完整的代码；请阅读本文，我这里的代码足够长，但不是全部:https://pastebin.com/NpspcJB8

作为我评估的一部分，我试图让我的代码将一个数组索引与另一个数组索引“同步”。出现的问题是，当我重复运行部分代码时，输​​入值后输出不会改变，例如:

case 'b':

    CaseBPrompt();

    for (t=0;t<350;t++) {
        int result2[350];
        result2[t] = strcmp(blankSpace, parkingSpace[t]); // determine if the value of parkingSpace is the same as strcmp "empty"
        Synchronise(t);

        if (result2[t] == 0) { // if value of parkingSpace == "empty"
            // use spaceNumber; sync int variable w/array index
            LevelPrint(t);

            strcpy(carCustomer[t], customerName);
            // copy value of car number to index
            strcpy(parkingSpace[t], carNumber); // omit array index to ensure this can run
            break; // required, as will print other available spaces otherwise
        }

    }
    printf("You will take %s's car\n", carCustomer[t]);
    printf("Please drive the car over to %c%d\n", level, spaceNumber[t][3]);
break;

t 是我在 main 外部全局声明的一个整型变量。以下是我这里用到的函数内容:

void CaseBPrompt() {
    printf("You have chosen to park a customer's car \n");
    printf("Please input the customer's name, using _ in lieu of a space\n");
    scanf(" %s", &customerName);
    // assign customer's name to carNumber - new array?

    printf("Please input the values of the number plate\n");
    // ensure use %s, as collecting string, and so const char/char error won't occur
    scanf(" %s", &carNumber);

}
void LevelPrint(int n) {
    if (spaceNumber[n][3] <= 100) {
    level = 'A';
    // don't forget to write out complete matrix for spaceNumber!

    printf("There is an available space at parking space %c%d\n", level,spaceNumber[n][3]);
    // assign parkingSpace array index w/user input

    } else if (spaceNumber[n][3] > 100 && spaceNumber[n][3]<= 200) {
        level = 'B';
        spaceNumber[n][3]-=100; // display correct parking space at corresponding level
        printf("There is an available space at parking space %c%d\n", level,spaceNumber[n][3]);

    } else if (spaceNumber[n][3] > 200 && spaceNumber[n][3]<= 300) {
        level = 'C';
        spaceNumber[n][3]-=200; // display correct parking space at corresponding level
        printf("There is an available space at parking space %c%d\n", level,spaceNumber[n][3]);

    } else if (spaceNumber[n][3] > 300 && spaceNumber[n][3]<= 350) {
        level = 'D';
        spaceNumber[n][3]-=300; // display correct parking space at corresponding level
        printf("There is an available space at parking space %c%d\n", level,spaceNumber[n][3]);

    } 
}
void Synchronise(int a) {
    spaceNumber[a][3] = n+1; // synchronise w/parking space
    carCustomer2[a][3] = n+1; // synchronise w/car customer name
}

当我运行代码时，我收到以下输出:

empty
350
25
Please write car number (without spaces):
b

b
Valet options (press the corresponding letter to select):
a) Check available spaces
b) Park a customer's car
c) Retrieve a customer's car
q) Exit the program
Your choice: b
You have chosen to park a customer's car
Please input the customer's name, using _ in lieu of a space
b
Please input the values of the number plate
b
There is an available space at parking space A1
You will take b's car
Please drive the car over to A1
Valet options (press the corresponding letter to select):
a) Check available spaces
b) Park a customer's car
c) Retrieve a customer's car
q) Exit the program
Your choice: b
You have chosen to park a customer's car
Please input the customer's name, using _ in lieu of a space
b
Please input the values of the number plate
b
There is an available space at parking space **A1**
You will take b's car
Please drive the car over to **A1**
Valet options (press the corresponding letter to select):
a) Check available spaces
b) Park a customer's car
c) Retrieve a customer's car
q) Exit the program
Your choice: c
You have chosen to retrieve a customer's car
Please write their name, again with _ in lieu of a space:
c
Please write down what was on their number plate:
90jda
c's car is at **1**.
Valet options (press the corresponding letter to select):
a) Check available spaces
b) Park a customer's car
c) Retrieve a customer's car
q) Exit the program
Your choice:

输出中用星号包围的字符是我所关心的，因为在 case 'c' 下添加代码之前，case b 下的代码会正确打印。我正在使用 DevC++，我想知道我的代码是否有问题。

当我再次选择执行选项 B 时，代码会为之前的每次迭代打印 A2、A3 等，并且选项 C 输出下显示的值是 149，因为我预先分配了一个值parkingSpace[150][3] 下的 90jda。

parkingSpace 下的值旨在与 case 'b' 的 blankSpace 同步，并打印出其中的可用空间。 for 循环。对于 case 'c'，parkingSpace 旨在与保存与用户输入 carNumber 中的字符串相同的任何索引进行同步，该字符串在case 'c' 下 for 循环的开头。

Answer 1

我相信我成功解决了这个问题，这是因为我在数组中声明的值之一不相同；程序没有返回错误，因为两者都被声明为整型变量