Question

所以在我的代码中，无论我改变了多少，我都无法让它正常工作它应该是有问题的。包括扫描对应于选项的int 然后它应该使用声明的选项调用导航并使用它但无论你选择什么选项，它都会说遗憾

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

#define OPENWINDOW "\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n"

void question(int option)

{
        printf("What Would You Like To Do?\n");
        printf("\t1.Add A Reminder\n\t2.View Reminders\n\t3.Manage Current Reminders\n\t4.Settings\n");
        scanf("%i", &option);

}

void navigate(int option)
{
    switch(option)
    {
        case 1:
        printf(OPENWINDOW);
        break;
        case 2:
        printf(OPENWINDOW);
        break;
        case 3:
        printf(OPENWINDOW);
        break;
        case 4:
        printf(OPENWINDOW);
        break;
        default :
        printf("sorry");
        question(option);


    }
}

int main()
{
     int option;    

     question(option);
     navigate(option);    

     return 0;
}

Answer 1

参数按值传递，而不是参考传递。因此，你的“选项”arg将在函数结束后很快“消失”。

如果将“引用”传递给var，则可以使用它来填充调用者变量。以下代码和示例修复了它。

void question(int *option)
{
        printf("What Would You Like To Do?\n");
        printf("\t1.Add A Reminder\n\t2.View Reminders\n\t3.Manage Current Reminders\n\t4.Settings\n");
        scanf("%i", option);
}

然后你这样称呼它：

int option;
question(&option);
// now you can use option...

由于函数可以返回值，您还可以：

int question(void)
{
        int option;
        printf("What Would You Like To Do?\n");
        printf("\t1.Add A Reminder\n\t2.View Reminders\n\t3.Manage Current Reminders\n\t4.Settings\n");
        scanf("%i", &option);
        return option;
}

// ...
int option = question();
// ...

导航和主要使用参考（指针）：

void navigate(int *option)
{
    switch(*option)
    {
        case 1:
          printf(OPENWINDOW);
          break;
        case 2:
          printf(OPENWINDOW);
          break;
        case 3:
          printf(OPENWINDOW);
          break;
        case 4:
          printf(OPENWINDOW);
          break;
        default:
          printf("sorry");
          question(option);    
    }
}

int main(void)
{
     int option;    

     question(&option);
     navigate(&option);    

     return 0;
}

Answer 2

您需要传递option作为传递引用。将选项的地址传递给question（）并在那里更新。

参考修改后的代码。

void question(int *option)
{
        printf("What Would You Like To Do?\n");
        printf("\t1.Add A Reminder\n\t2.View Reminders\n\t3.Manage Current Reminders\n\t4.Settings\n");
        scanf("%i", option);
}

将问题（）称为，

question(&option);

Answer 3

您正在通过值option

传递变量question(option)

您应该通过引用传递option变量

void question(int *option)
{
        printf("What Would You Like To Do?\n");
        printf("\t1.Add A Reminder\n\t2.View Reminders\n\t3.Manage Current Reminders\n\t4.Settings\n");
        scanf("%i", option);

}

void navigate(int *option)
{
    switch(*option)
    {
        case 1:
        printf(OPENWINDOW);
        break;
        case 2:
        printf(OPENWINDOW);
        break;
        case 3:
        printf(OPENWINDOW);
        break;
        case 4:
        printf(OPENWINDOW);
        break;
        default :
        printf("sorry");
        question(option);

}

int main()
{
     int option;    

     question(&option);
     navigate(&option);    

     return 0;
}

有关此问题的详细信息，请查看此链接Difference between call by reference and call by value

Answer 4

您正在传递“选项”作为按值调用。因此无论你传递给问题（）。会丢失。

或者，你从question（）返回“option”并将其传递给navigate（）。

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

#define OPENWINDOW "\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n"

int question()
{       int option;
        printf("What Would You Like To Do?\n");
        printf("\t1.Add A Reminder\n\t2.View Reminders\n\t3.Manage Current Reminders\n\t4.Settings\n");
        scanf("%i", &option);
        return option;

}

void navigate(int option)
{
    switch(option)
    {
        case 1:
        printf(OPENWINDOW);
        break;
        case 2:
        printf(OPENWINDOW);
        break;
        case 3:
        printf(OPENWINDOW);
        break;
        case 4:
        printf(OPENWINDOW);
        break;
        default :
        printf("sorry");
        question(option);


    }
}

int main()
{
int option;

option = question();
navigate(option);

return 0;
}
~

Answer 5

您需要将option的指针传递给question或从函数question返回。

在您的情况下，option中main()的值question()在int question() { int option; printf("What Would You Like To Do?\n"); printf("\t1.Add A Reminder\n\t2.View Reminders\n\t3.Manage Current Reminders\n\t4.Settings\n"); scanf("%i", &option); return option; } int main() { int option; option = question(option); navigate(option); return 0; }中读取时未发生变化。将代码更新为

{{1}}

Answer 6

如果您不想使用pass-by-reference，则可以使用在代码中使用的pass-by-value。它只需要正确实施。你可以改变你的问题＆＃34;通过将void更改为＆＃34; int＆＃34;来返回值并在问题结束前发出一份退货声明。检查以下代码：

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

#define OPENWINDOW "\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n"

int question()

{
        printf("What Would You Like To Do?\n");
        printf("\t1.Add A Reminder\n\t2.View Reminders\n\t3.Manage Current Reminders\n\t4.Settings\n");
        scanf("%i", &option);
        return i;
}

void navigate(int option)
{
    switch(option)
    {
        case 1:
        printf(OPENWINDOW);
        break;
        case 2:
        printf(OPENWINDOW);
        break;
        case 3:
        printf(OPENWINDOW);
        break;
        case 4:
        printf(OPENWINDOW);
        break;
        default :
        printf("sorry");
        question(option);
    }
}

int main()
{
int option;
option = question(option);
navigate(option);

return 0;
}

Answer 7

因为变量选项只将其值传递给函数question（），所以变量选项的值确实不变，所以，也许你应该在函数问题中返回option的值（）

C / C ++函数没有正确调用开关结构

7 个答案: