c - %lf' 需要类型为 'double' 的参数，但参数 2 的类型为 'double *'

Question

为什么我在尝试编译我的程序时收到此警告？%lf' 需要“double”类型的参数，但参数 2 的类型为“double *”我正在使用 CodeBlocks IDE，但这些行给出了一个巨大的数字:

double calculate1 = mealPrice * (double)(percentage)/100;
printf("The tip that you should leave is %lf \n", &calculate1);

我是 C 编程的新手，仍在学习东西。

// CS 262, Lab Section <208>
// Lab 2

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

int main(){
   printf("Enter the price of the meal: \n");
   double mealPrice = 0;
   scanf("%lf\n", &mealPrice);

   printf("Now enter the tip percentage: \n");
   int percentage = 0;
   scanf("%d\n", &percentage);


   //Calculates tip amount in double, int, and float types
   double calculate1 = mealPrice * (double)(percentage)/100;
   printf("The tip that you should leave is %lf \n", &calculate1);
   int calculate2 = (int)mealPrice * (int)(percentage/100);
   printf("The tip that you should leave is %d\n", &calculate2);
   float calculate3 = (float)mealPrice * (float)(percentage/100);
   printf("The tip that you should leave is &f\n", &calculate3);

   //Add tip to meal price
   double total = calculate1 + mealPrice;
   printf("The total price including tips is %lf\n", total);

   printf("The meal cost is %f\nThe tip percentage is %d\nThe tip amount is%lf\nThe total cost is %lf\n", &mealPrice, &percentage, &calculate1, &total);
   return 0;
}

Answer 1

问题是您不应该将指针与 printf 一起使用(除非您正在使用指针格式说明符)。您通常通过指向 scanf 的指针传递事物(因为它需要更改它们)并通过值传递给 printf(因为它不应该更改它们)。这就是为什么你会得到巨大的数字。

它应该是这样的:

printf("The tip that you should leave is %lf \n", calculate1);

和

scanf("%lf\n", &mealPrice);

不是

printf("The tip that you should leave is %lf \n", &calculate1);

或

scanf("%lf\n", mealPrice);

此外，以后不要向我们显示编译器警告，除非它们与您发布的代码特别相关，否则您会得到困惑的响应。