我正在写一个程序，你有一定数量的钱，你可以用这些钱购买柠檬水袋来制作柠檬水。我有一个功能：
void buy_lemons(double *pLemons, double *pMoney);
每当我从main调用它时，都会给我这个错误：
error: incompatible type for argument 1 of 'buy_lemons'
在main之前的函数原型中有另一个注释：
note: expected 'double *' but argument is of type 'double'
到目前为止，这是我的代码：
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
// Symbolic constants to be used.
// Prices of bags of sugar and lemon, respectively.
#define PRICE_LEMON 3.50
#define PRICE_SUGAR 2.00
// Fraction of a bag of lemons and sugar used on a single cup of lemonade.
#define LEMON_PER_CUP 0.03
#define SUGAR_PER_CUP 0.04
// The initial loan the user is given to start their lemonade stand.
#define START_MONEY 20.00
// Using symbolic constants for true and false.
#define FALSE 0
#define TRUE 1
// Function prototypes - do not change these
void buy_lemons(double *pLemons, double *pMoney);
void buy_sugar(double *pSugar, double *pMoney);
// Main function
int main() {
int num_day, ans, climate, cost;
double money = START_MONEY, num_lemons = 0.00, num_sugar = 0.00;
srand(time(0));
printf("Welcome to the Game of Lemonade!\n\n");
printf("You start the game with $%.2lf and no supplies!\n", START_MONEY);
// Loop through each day. Ask the user if they want to buy lemons. If so,
// carry out the transaction. Then ask them if they want to buy sugar.
// If so, do this transaction as well. Then, let them sell lemonade for
// the day. Finally, print a status report after they've sold lemonade
// at the end of the day.
for (num_day = 1; num_day <= 10; num_day++) {
printf("Would you like to buy some lemons? (1 - Yes, 0 - No)\n");
scanf("%d", &ans);
if(ans == 1)
buy_lemons(num_lemons, money);
printf("1a\n");
if(ans == 0)
printf("2a\n");
printf("Would you like to buy some sugar? (1 - Yes, 0 - No)\n");
scanf("%d", &ans);
if(ans == 1)
buy_sugar(num_sugar, money);
printf("1b\n");
if(ans == 0)
printf("2b\n");
}
return 0;
}
// Pre-conditions: pLemons and pMoney are pointers to variables that store
// the user's number of bags of lemons left and amount of
// money left.
// Post-condition: The user is given the opportunity to buy lemons. If
// successful, the number of bags of lemons and the amount
// of money the user has are adjusted accordingly.
//
// What to do in this function: If the user doesn't have enough money to
// even buy one bag of lemons, tell them so and return. Otherwise, ask
// the user how many bags of lemons they want to buy. If they answer less
// than one, tell them they must get more and reprompt them. If they
// answer more than they can buy, tell them they don't have that much
// money and reprompt them. Continue prompting them until they answer with
// a valid value. Then process the transaction.
void buy_lemons(double *pLemons, double *pMoney) {
double bags, total;
if (*pMoney < PRICE_LEMON)
printf("You do not have enough money to buy any lemons.\n");
if(*pMoney >= PRICE_LEMON){
printf("How many bags of lemons would you like to purchase?\n");
scanf("%.2lf\n", &bags);
while (bags < 1){
printf("You must buy at least 1 bag of lemons.\n");
printf("How many bags of lemons would you like to purchase?\n");
scanf("%.2lf\n", &bags);
}
total = PRICE_LEMON * bags;
printf("The total is %lf.\n", &total);
while(total > *pMoney){
printf("You don't have enough money.\n");
printf("How many bags of lemons would you like to purchase?\n");
scanf("%.2lf\n", &bags);
total = PRICE_LEMON * bags;
}
*pLemons += bags;
*pMoney -= total;
}
}
// Pre-conditions: pSugar and pMoney are pointers to variables that store
// the user's number of bags of lemons left and amount of
// money left.
// Post-condition: The user is given the opportunity to buy sugar. If
// successful, the number of bags of sugar and the amount
// of money the user has are adjusted accordingly.
//
// What to do in this function: If the user doesn't have enough money to
// even buy one bag of sugar, tell them so and return. Otherwise, ask
// the user how many bags of sugar they want to buy. If they answer less
// than one, tell them they must get more and reprompt them. If they
// answer more than they can buy, tell them they don't have that much
// money and reprompt them. Continue prompting them until they answer with
// a valid value. Then process the transaction.
void buy_sugar(double *pSugar, double *pMoney) {
double bags, total;
if (*pMoney < PRICE_SUGAR)
printf("You do not have enough money to buy any sugar.\n");
if(*pMoney >= PRICE_SUGAR){
printf("How many bags of sugar would you like to purchase?\n");
scanf("%.2lf\n", &bags);
while (bags < 1){
printf("You must buy at least 1 bag of sugar.\n");
printf("How many bags of sugar would you like to purchase?\n");
scanf("%.2lf\n", &bags);
}
total = PRICE_SUGAR * bags;
printf("The total is %lf.\n", &total);
while(total > *pMoney){
printf("You don't have enough money.\n");
printf("How many bags of sugar would you like to purchase?\n");
scanf("%.2lf\n", &bags);
total = PRICE_SUGAR * bags;
}
*pSugar += bags;
*pMoney -= total;
}
}
有什么想法吗？
答案 0 :(得分：3)
而不是
buy_lemons(num_lemons, money);
尝试
buy_lemons(&num_lemons, &money);
推理：＆amp;放在变量名称之前的是'address-of'运算符。如果将地址传递给整数而不是数字值，则可以将其解释为指针并取消引用，以便即使在存储它的原始位置也可以编辑该值。
e.g。如果num_lemons的类型为double，则&num_lemons的类型为double*
P.S。这就是为什么你把＆amp;在非指针参数传递给scanf