我正在尝试编写一个C ++代码，用户输入一个日期，然后使用John Conway的Doomsday算法输出该日期的日期。
我正在尝试访问在函数dayofthemonth中声明和定义的变量z，并在名为dayoftheweek的函数中使用它。请记住，我对C ++很陌生，所以如果你能尽可能简单地回答，那将会有很大帮助。这是代码：
#include <iostream>
#include <string>
using namespace std;
int dayofthemonth (int year, int month){
int z;
if ((year % 400 == 0 || year % 100 !=0) && (year % 4 == 0)){ //reference day of the month for leap years
cout << year << " is a leap year." << endl;
if (month == 1){
cout << "The doomsday for month " << month << " is 4." << endl;
z = 4;
}
if (month == 2){
cout << "The doomsday for month " << month << " is 1." << endl;
z = 1;
}
if (month == 3){
cout << "The doomsday for month " << month << " is 0." << endl;
z = 0;
}
if (month == 4){
cout << "The doomsday for month " << month << " is 4." << endl;
z = 4;
}
if (month == 5){
cout << "The doomsday for month " << month << " is 9." << endl;
z = 9;
}
if (month == 6){
cout << "The doomsday for month " << month << " is 6." << endl;
z = 6;
}
if (month == 7){
cout << "The doomsday for month " << month << " is 11." << endl;
z = 11;
}
if (month == 8){
cout << "The doomsday for month " << month << " is 8." << endl;
z = 8;
}
if (month == 9){
cout << "The doomsday for month " << month << " is 5." << endl;
z = 5;
}
if (month == 10){
cout << "The doomsday for month " << month << " is 10." << endl;
z = 10;
}
if (month == 11){
cout << "The doomsday for month " << month << " is 7." << endl;
z = 7;
}
if (month == 12){
cout << "The doomsday for month " << month << " is 12." << endl;
z = 12;
}
}
else{ //reference day of the month for non-leap years
cout << year << " is not a leap year." << endl;
if (month == 1){
cout << "The doomsday for month " << month << " is 3." << endl;
z = 3;
}
if (month == 2){
cout << "The doomsday for month " << month << " is 0." << endl;
z = 0;
}
if (month == 3){
cout << "The doomsday for month " << month << " is 0." << endl;
z = 0;
}
if (month == 4){
cout << "The doomsday for month " << month << " is 4." << endl;
z = 4;
}
if (month == 5){
cout << "The doomsday for month " << month << " is 9." << endl;
z = 9;
}
if (month == 6){
cout << "The doomsday for month " << month << " is 6." << endl;
z = 6;
}
if (month == 7){
cout << "The doomsday for month " << month << " is 11." << endl;
z = 11;
}
if (month == 8){
cout << "The doomsday for month " << month << " is 8." << endl;
z = 8;
}
if (month == 9){
cout << "The doomsday for month " << month << " is 5." << endl;
z = 5;
}
if (month == 10){
cout << "The doomsday for month " << month << " is 10." << endl;
z = 10;
}
if (month == 11){
cout << "The doomsday for month " << month << " is 7." << endl;
z = 7;
}
if (month == 12){
cout << "The doomsday for month " << month << " is 12." << endl;
z = 12;
}
}
}
int doomsday (int year){ //reference day of the week
int a, b, c , d, e, x;
a = ((year/100) % 4);
b = (year % 100);
c = (b/12);
d = (b % 12);
e = (d/4);
x = ((c + d + e + (5*a) + 2) % 7);
if (x == 0){
cout << "The doomsday for " << year << " is Sunday." << endl;
}
else if (x == 1){
cout << "The doomsday for " << year << " is Monday." << endl;
}
else if (x == 2){
cout << "The doomsday for " << year << " is Tuesday." << endl;
}
else if (x == 3){
cout << "The doomsday for " << year << " is Wednesday." << endl;
}
else if (x == 4){
cout << "The doomsday for " << year << " is Thursday." << endl;
}
else if (x == 5){
cout << "The doomsday for " << year << " is Friday." << endl;
}
else if (x == 6){
cout << "The doomsday for " << year << " is Saturday." << endl;
}
}
void dayoftheweek(int month, int day, int year}(
int r;
cout << "You want to find out what day of the week " << month << "/" << day << "/" << year << " lies on." << endl;
doomsday(year);
r = (day - z) + 14; //offset between the given day and the result of dayofthemonth function.
cout << r << endl;
}
int main(){
int year, month, day;
cout << "Enter the year." << endl;
cin >> year;
cout << "Enter the month using numbers 1-12." << endl;
cin >> month;
cout << "Enter the day." << endl;
cin >> day;
dayoftheweek(month, day, year);
system ("PAUSE");
return 0;
}
答案 0 :(得分：1)
您的函数的返回类型为int，但实际上并未返回任何值。返回z，然后将其作为参数传递给下一个函数。
答案 1 :(得分：1)
另一个观察结果，此函数int dayofthemonth (int year, int month)没有调用者。
可能你应该改为，
r = (day - z) + 14;到
r = (day - dayofthemonth(year, month)) + 14; //如果此函数返回z。
答案 2 :(得分：0)
如果您想要一种简单的方法在代码的所有范围中使用一个变量，您可以在main函数之前和“#include＆lt; ...＆gt;”之后和“using namespace ...”之后声明它。 但是当你使用这种方式时要小心，因为如果你在代码中使用该变量2次或更多次并且不使它为0，你将看到错误的答案。 最后我在我的代码“system（”pause“）中使用了”cin.get（）“，因为如果你想在linux中编译你的代码”system（“pause”）“会给你一个错误。< / p>
#include <iostream>
#include <string>
using namespace std;
int z;
void dayofthemonth (int year, int month)
{
z = 0;
if ((year % 400 == 0 || year % 100 !=0) && (year % 4 == 0)){ //reference day of the month for leap years
cout << year << " is a leap year." << endl;
if (month == 1){
cout << "The doomsday for month " << month << " is 4." << endl;
z = 4;
}
if (month == 2){
cout << "The doomsday for month " << month << " is 1." << endl;
z = 1;
}
if (month == 3){
cout << "The doomsday for month " << month << " is 0." << endl;
z = 0;
}
if (month == 4){
cout << "The doomsday for month " << month << " is 4." << endl;
z = 4;
}
if (month == 5){
cout << "The doomsday for month " << month << " is 9." << endl;
z = 9;
}
if (month == 6){
cout << "The doomsday for month " << month << " is 6." << endl;
z = 6;
}
if (month == 7){
cout << "The doomsday for month " << month << " is 11." << endl;
z = 11;
}
if (month == 8){
cout << "The doomsday for month " << month << " is 8." << endl;
z = 8;
}
if (month == 9){
cout << "The doomsday for month " << month << " is 5." << endl;
z = 5;
}
if (month == 10){
cout << "The doomsday for month " << month << " is 10." << endl;
z = 10;
}
if (month == 11){
cout << "The doomsday for month " << month << " is 7." << endl;
z = 7;
}
if (month == 12){
cout << "The doomsday for month " << month << " is 12." << endl;
z = 12;
}
}
else{ //reference day of the month for non-leap years
cout << year << " is not a leap year." << endl;
if (month == 1){
cout << "The doomsday for month " << month << " is 3." << endl;
z = 3;
}
if(month == 2){
cout << "The doomsday for month " << month << " is 0." << endl;
z = 0;
}
if (month == 3){
cout << "The doomsday for month " << month << " is 0." << endl;
z = 0;
}
if (month == 4){
cout << "The doomsday for month " << month << " is 4." << endl;
z = 4;
}
if (month == 5){
cout << "The doomsday for month " << month << " is 9." << endl;
z = 9;
}
if (month == 6){
cout << "The doomsday for month " << month << " is 6." << endl;
z = 6;
}
if (month == 7){
cout << "The doomsday for month " << month << " is 11." << endl;
z = 11;
}
if (month == 8){
cout << "The doomsday for month " << month << " is 8." << endl;
z = 8;
}
if (month == 9){
cout << "The doomsday for month " << month << " is 5." << endl;
z = 5;
}
if (month == 10){
cout << "The doomsday for month " << month << " is 10." << endl;
z = 10;
}
if (month == 11){
cout << "The doomsday for month " << month << " is 7." << endl;
z = 7;
}
if (month == 12){
cout << "The doomsday for month " << month << " is 12." << endl;
z = 12;
}
}
}
void doomsday (int year)
{ //reference day of the week
int a, b, c , d, e, x;
a = ((year/100) % 4);
b = (year % 100);
c = (b/12);
d = (b % 12);
e = (d/4);
x = ((c + d + e + (5*a) + 2) % 7);
if (x == 0)
cout << "The doomsday for " << year << " is Sunday." << endl;
else if (x == 1)
cout << "The doomsday for " << year << " is Monday." << endl;
else if (x == 2)
cout << "The doomsday for " << year << " is Tuesday." << endl;
else if (x == 3)
cout << "The doomsday for " << year << " is Wednesday." << endl;
else if (x == 4)
cout << "The doomsday for " << year << " is Thursday." << endl;
else if (x == 5)
cout << "The doomsday for " << year << " is Friday." << endl;
else if (x == 6)
cout << "The doomsday for " << year << " is Saturday." << endl;
}
void dayoftheweek(int month, int day, int year)
{
int r;
cout << "You want to find out what day of the week " << month << "/" << day << "/" << year << " lies on." << endl;
doomsday(year);
r = (day - z) + 14; //offset between the given day and the result of dayofthemonth function.
cout << r << endl;
}
int main(){
int year, month, day;
cout << "Enter the year." << endl;
cin >> year;
cout << "Enter the month using numbers 1-12." << endl;
cin >> month;
cout << "Enter the day." << endl;
cin >> day;
dayoftheweek(month, day, year);
cin.get();
return 0;
}