Question

我正在尝试创建一个存储不同值的结构并通过指针引用它们，并在函数中使用该指针，但是一旦我将指针传递给函数，它似乎丢失了原始数据中的所有数据结构指针也指向了。

这是我的代码：

    #include <stdio.h>
    #include <stdlib.h>
    #include <Windows.h>
    #pragma warning(disable:4996)

     void clear_screen();
     void stopE();
     void cost(struct coffee *cust1);
     void reciept(char name[], int coff, int tea, double total, double noTax);
     double noTax(struct coffee *cust1);
    struct coffee
    {
        char name[21];
        int cCups;
        int tCups;
        double totalCost;
    };

    void main()
    {
        char buffer[10] = {0};
        struct coffee cust;

        printf("Please input the customers name: ");
        gets(cust.name);
        printf(" \nPlease enter how many cups of coffee ordered: ");
        gets(buffer);
        cust.cCups = atoi(buffer);
        stopE();
        printf(" \nPlease enter how many cups of tea ordered: ");
        gets(buffer);
        cust.tCups = atoi(buffer);
        stopE();
        struct coffee *cust1 = &cust;


        clear_screen();
        double withoutTax = noTax(&cust1);
        cost(&cust1);
        reciept(cust.name, cust.cCups, cust.tCups, cust.totalCost, withoutTax);


    }

    void stopE()
    {
        int c = getchar();
        while (c != EOF && c != '\n')
        {
            c = getchar();
        }
    }
    void cost(struct coffee *cust1)
    {
        double a = (double)cust1->cCups*1.75;
        double b = (double)cust1->tCups * 1.5;

        double inter = a + b;
        double totalCost = inter + (inter*0.13);
        cust1->totalCost = totalCost;
    }
    double noTax(struct coffee *cust1)
    {
        double totalCost = (cust1->cCups*1.75) + (cust1->tCups*1.5);
        return totalCost;
    }
    void reciept(char name[], int coff, int tea, double total, double noTax)
    {
        printf("Coffee Co.\n");
        srand(time(NULL));
        int r = rand() % 101;
        printf("Order Number: %d\n", r);
        printf("Name: %s\n", name);

        for (int i = 0; i < coff; i++)
        {
            printf("Coffee:            $1.75\n");
        }

        for (int i = 0; i < tea; i++)
        {
            printf("Tea:               $1.50\n");
        }
        printf("Total without tax: $%.2f", noTax);
        printf("\n13%% Tax total:     $%.2f", (noTax*0.13));
        p

rintf("\nTotal:             $%.2f\n", total);
    getchar();

}
void clear_screen(void)//Function to clear screen 
{
    DWORD n;                         /* Number of characters written */
    DWORD size;                      /* number of visible characters */
    COORD coord = { 0 };               /* Top left screen position */
    CONSOLE_SCREEN_BUFFER_INFO csbi;

    /* Get a handle to the console */
    HANDLE h = GetStdHandle(STD_OUTPUT_HANDLE);

    GetConsoleScreenBufferInfo(h, &csbi);

    /* Find the number of characters to overwrite */
    size = csbi.dwSize.X * csbi.dwSize.Y;

    /* Overwrite the screen buffer with whitespace */
    FillConsoleOutputCharacter(h, TEXT(' '), size, coord, &n);
    GetConsoleScreenBufferInfo(h, &csbi);
    FillConsoleOutputAttribute(h, csbi.wAttributes, size, coord, &n);

    /* Reset the cursor to the top left position */
    SetConsoleCursorPosition(h, coord);
}

如果有人可以帮助我，我将不胜感激，我无法弄清楚如何将数据传递给函数。

编辑：该编译指示是使用scanf删除错误，阻止我运行我的程序，因为它不安全。

Answer 1

你太努力了：

替换：

struct coffee *cust1 = &cust;

clear_screen();
double withoutTax = noTax(&cust1);
cost(&cust1);

由：

clear_screen();
double withoutTax = noTax(&cust);
cost(&cust);

&cust是cust的地址，换句话说：它是指向curst的指针。

你没有得到编译器警告吗？

Answer 2

该行

struct coffee *cust1 = &cust;

是不必要的。我看到当你将这个指针传递给你的定价函数时，你会用

这样做

double withoutTax = noTax(&cust1); cost(&cust1);

这是将（＆amp;）指针cust1的地址传递给函数。你想要做的只是传递指针，或者更简单地传递原始结构的地址。

double withoutTax = noTax(cust1); cost(cust1);或

double withoutTax = noTax(&cust); cost(&cust);

我在一个在线IDE中对该程序进行了测试，看起来这个改变似乎正常。

指向从原始结构中丢失数据的结构的指针

2 个答案: