c - c中函数的声明和使用

Question

我找不到这段代码中的错误。我正在尝试编写一个程序，根据日、月和年返回一年中的天数(例如:60)[例如:1/3/2000(2000 年 3 月 3 日)]。

编译器给我这些错误:

• too few arguments to function 'day_of_year'
• conflicting types for 'day_of_year'

#include <stdio.h>

int day_of_year(int day, int month, int year);

int main(){
    int day, month, year, i, count=0;
    int a[]={31,28,31,30,31,30,31,31,30,31,30,31};
    printf ("Enter the day: ");
    scanf ("%d", &day);
    printf ("Enter the month: ");
    scanf ("%d", &month);
    printf ("Enter the year: ");
    scanf ("%d", &year);
    count=day_of_year();
    printf ("Count: %d", count);
    return 0;
}

int day_of_year (int day, int month, int year, int i, int count){
    int a[]={31,28,31,30,31,30,31,31,30,31,30,31};
    if (year%4==0) a[2]++;
    count = day;
    for (i=0;i<month;i++)
        count+=a[i];
    return count;
}

Answer 1

声明:

int day_of_year(int day, int month, int year);

调用:

count=day_of_year();

定义:

int day_of_year (int day, int month, int year, int i, int count){

C 是一种强类型语言，这意味着函数参数的数量和类型非常匹配。

因此，您需要更正参数列表并将缺少的参数添加到函数调用中。

看起来您并不真的需要定义中的最后两个参数 - 而是将它们声明为局部变量:

int day_of_year (int day, int month, int year){
    int i, count;