二进制 / 的无效操作数（有“float *”和“int”）？

Question

我需要通过引用传递我的班级。我在标题中遇到了错误，但仅限于功能 divideByTweleve 和 multiplyByEleven。这是为什么？我究竟做错了什么？ 请帮我解决这个问题。

#include <stdio.h>
void displayMenu();
float addTen (float* number);
float divideByTwelve(float* number);
float subtractSixteen(float* number);
float multiplyByEleven(float* number);
int menu;
float number, finalNumber;
float* ptr_number;

int main (void)
{
    float* ptr_number=NULL;
    ptr_number=&number;

    printf("Please enter a number : ");
    scanf("%f", &number);
    displayMenu();
    scanf("%d", &menu);

do {
    printf("\nEnter 5 to see final number\n");
    scanf("%d",&menu);
    switch (menu)
        {
        case 1: number=addTen(ptr_number); 
                break; 
        case 2: number=divideByTwelve(ptr_number);
                break; 
        case 3: number=subtractSixteen(ptr_number);
                break;
        case 4: number=multiplyByEleven(ptr_number);
                break;
        }
    printf("Your number is: %.2f \n", number);
    }while(menu!=5);//close of the do-while loop
}

void displayMenu()
{
    printf("How would you like to manipulate your number\n");
    printf("1. Add 10\n");
    printf("2. Divide by 12\n");
    printf("3. Subtract 16\n");
    printf("4. Multiply by 11\n");
}


float addTen (float* number)
{
    ptr_number=ptr_number+10;

}

float divideByTwelve (float* number)
{
    ptr_number=ptr_number/12;

}

float subtractSixteen(float* number)
{
    ptr_number=ptr_number-16;

}

float multiplyByEleven(float* number)
{
    ptr_number=ptr_number*11;

}

Answer 1

这是您的（许多）问题之一。

您将addTen 声明为：

float addTen (float* number);

并将其称为：

number=addTen(ptr_number); 

这清楚地表明它返回一个值。

现在，您在函数定义中看到返回语句了吗？

float addTen (float* number)
{
    ptr_number=ptr_number+10;   // No return statement!!
}

---

接下来，您的所有函数都将float* 作为参数，您应该将其读作“指向浮点数的指针”。请注意，指针与值本身不是。

要从指向值的指针中获取，您需要“取消引用”它：

float addTen (float* number)
{
    *ptr_number = *ptr_number + 10;
}

每个ptr_number前面的星号（*）表示从指针返回到实际值。

之前，您试图将 10 添加到指针。
（这是有效的，但在当前上下文中不合适）

现在，使用*，您将存储在指向位置的值加10。

---

---

最后，这是我修改后的代码版本

void displayMenu();
void addTen (float* number);
void divideByTwelve(float* number);
void subtractSixteen(float* number);
void multiplyByEleven(float* number);


int main (void)
{
    int menu;
    float number;
    
    printf("Please enter a number : ");
    scanf("%f", &number);
    displayMenu();
    scanf("%d", &menu);

    do {
    printf("\nEnter 5 to see final number\n");
    scanf("%d",&menu);
    switch (menu)
        {
        case 1: addTen(&number);             break; 
        case 2: divideByTwelve(&number);     break; 
        case 3: subtractSixteen(&number);    break;
        case 4: multiplyByEleven(&number);   break;
        }
    printf("Your number is: %.2f \n", number);
    }while(menu!=5);//close of the do-while loop

    printf("Your FINAL number is: %.2f \n", number);
}

void displayMenu()
{
    printf("How would you like to manipulate your number\n");
    printf("1. Add 10\n");
    printf("2. Divide by 12\n");
    printf("3. Subtract 16\n");
    printf("4. Multiply by 11\n");
}


void addTen (float* ptr_number)
{
    *ptr_number = *ptr_number+10;
}

void divideByTwelve (float* ptr_number)
{
    *ptr_number = *ptr_number/12;
}

void subtractSixteen(float* ptr_number)
{
    *ptr_number = *ptr_number-16;
}

void multiplyByEleven(float* ptr_number)
{
    *ptr_number = *ptr_number*11;
}

Answer 2

因为您没有从这些函数返回 float 值。如果你想传递一个指向你的函数的指针，那么你应该这样做

float divideByTwelve (float *)
{
    *ptr_number= *ptr_number/12;
     return *ptr_number;
}

但是您已将number 声明为全局变量，因此无需将指针传递给这些函数。只需传递number 本身`;

float divideByTwelve (float number)
{
     return number / 12;
}