Hello i been doing some program , my program is to get the number of days passed in the year. Now when i try to run and the output gives me "4438232".
For example if the user enter (mm-dd-yy) 3-18-2013, then the reaming days passed in the year is 77.
Does this code make sense on getting the reaming days passed in the year?
void dateType::Num_DayPassed()
{
int sum;
int yy = 365;
if (month ==1)
{
cout<<"Number of days Passed in the Year: "<<sum<<endl;
day=31;
sum=day-yy;
}
........
continued until month 12..
Full Code
Output
C++ programs run top-to-bottom. You're outputting sum before you set it.
Related
im writing a simple program that you give a number of days and it gives back the number of years weeks and days that equal to the numbers of days you give.
but i noticed that you can get two different answers even tho when i checked the math it makes sense in both cases .
can someone please please explain to me why the answers are different and which one is correct
#include<iostream>
using namespace std;
int main()
{
int y;
int d,w;
int Days;
cin>>d;
y=d/365;
int LessThanAYearDays = d%365;
Days=LessThanAYearDays%7;
w=LessThanAYearDays/7;
int SameDays = d%7;
cout<<"answer1 is : "<<y<<" "<<w<<" "<<Days<< "\n";
cout<<"answer2 is : "<<y<<" "<<w<<" "<<SameDays<< "\n";
return 0;
}
There are not an exact multiple of weeks in a year, so you are displaying different things. They will be the same in years divisible by 7.
E.g. Assume that a given year starts on a Tuesday.
Days corresponds to how many days past the last Tuesday you are.
SameDays corresponds to what day you are on.
See also the distinction that std makes between a std::chrono::duration, which is a count of time, and a std::chrono::time_point, which is a count of time since a particular date.
Alright, so I had an assignment in c++ where the user was to input a day and if the day was less than 10, the program would have to output a 0 in front of it. If the day was greater than 10, the program would output the day. An example would be if the user inputs 5, the program would output 05. If the user inputs 24, the program would output 24. My question is how do we define the 0 in front of the 5. This is the snippet of code of which I attempted but to no avail.
if (day < 10){
f = 0 << day;
}
else {
f = day;
}
This is the statement I'm struggling with: f = 0 << day;
Any help would be great.
Thanks.
You can simply write:-
if(day<10)
{
cout<<"0"<<day;
}
else cout<<day;
If you have any queries, please let me know
Use printf to format your output
Ex:
printf("%02d", day);
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I have a program that is getting a date from some protocol, in format of DD/MM/YYYY.
The Problem is that I need to know the info of that day(day of the week, day in the year...) and I don't know how to do it.
Usually, when I want to get a day info, I am using time(time_t*) and convert the result to tm struct using localltime_r(tm*, time_t*) and then I have everything i need.
But in this case, this is not the current time(so I can not use time(time_t*)) and I don't have nothing except the date.
I can create a new tm struct and fill only tm_year, tm_mon, tm_mday and use mktime(tm*), but I am not sure if this will give me the right detail's of the desired date.
You might consider using Howard Hinnant's free, open-source date/time library. It works with C++11 and later, ported to gcc, clang and VS. It has been accepted into the C++20 draft specification.
Example code:
#include "date/date.h"
#include <iostream>
#include <sstream>
int
main()
{
using namespace std;
using namespace date;
istringstream in{"09/07/2018"};
sys_days sd;
in >> parse("%d/%m/%Y", sd);
cout << "Day of the week is " << weekday{sd} << '\n';
cout << "Day of the year is " << sd - sys_days{year_month_day{sd}.year()/1/0} << '\n';
}
Output:
Day of the week is Mon
Day of the year is 190d
If you would rather do the computations yourself, here are extremely efficient non-iterative public domain algorithms for calendrical computations. The library referenced above is nothing but a type-safe wrapper around these algorithms with more pleasant syntax.
In case you wanted to get write methods which get the day of week/year as opposed to using a library, here's what I would do.
You're going to need to take leap years into account. For any given year, determine whether or not it is a leap year. Something like this would work (using a formula found here: https://en.wikipedia.org/wiki/Leap_year#Algorithm:
bool isLeapYear(short year)
{
bool leapYear = false;
if (((year % 4 == 0 && year % 100 != 0)) || (year % 400 == 0))
leapYear = true;
return leapYear;
}
From this, calculating the day of the year is straightforward. Simply add up all the days for each month, if it is a leap year, add 29 days onto your tally for day of year if you come across February.
As for finding the day of the week, it really helps if you start with some lower bound for the year (in this case LOWYEAR = 1760) and start with the first day of that year (STARTDAYOFWEEK = 2). Each day of the week is corresponds to a number (0-6) where Sunday is 0, Monday is 1, etc.
int DayOfWeek(void)
{
//jan 1 1760 was tuesday (2)
short dayCount = STARTDAYOFWEEK;
for (unsigned i = LOWYEAR; i < year; i++)
{
if (isLeapYear(i))
dayCount += 2;
else
dayCount++;
}
return (dayCount + dayOfYear) - 1) % DAYSINWEEK;
}
Finding the day of week is now really easy, after calculating dayCount, the day of week is found by adding dayCount with dayOfYear and modding by DAYSINWEEK (6).
The resulting number will correspond to the day of week (0-6).
I am trying to create a program that finds out how many months they have been alive, but have been running into some issues. Here is my function so far:
int getResult(int year, month, day, endResult)
{
int thisYear, thisMonth, thisDay;
year = thisYear - year;
year *= 12;
}
And what I'm trying to accomplish would show an output like:
Output:
What year were you born?
1989
What month were you born?
5
What day were you born?
23
You are x months old.
I was going to continue with months but then I realized, what if the month they were born is in after this month or before? So, if anyone has any tips on how to calculate that, I'd appreciate it.
Let's see. First, let's say now is:
year_now and month_now
and your birthday is:
year_birth and month_birth
Now, we go case by case:
month_now == month_birth: as you have already computed:
months_old = (year_now-year_birth)*12
month_now > month_birth: easily, you have:
months_old = (year_now-year_birth)*12 + (month_now-month_birth)
month_now < month_birth: in this case, (year_now-year_birth)*12 gives you more months then necessary, and you have to subtract:
months_old = (year_now-year_birth)*12 - (month_birth-month_now)
Now if you look carefully, you will see that they are all in fact the same formula:
months_old = (year_now-year_birth)*12 + (month_now-month_birth)
(in the third case, month_now-month_birth is negative)
months = (thisyear-years)*12+(thisMonth-months)
if(months < 0)
System.out.println("Invalid info")
else{
//DO YOUR THANG BRO
}
Hi to all thank all in advance to those who tried to answer or answer and part of this question.
Calculate the sum of the digits of the year.
Calculate the absolute value of the difference between the year and the ’reverse’ of the year.
Calculate the number of even factors of the day.
Calculate the greatest common divisor of the day, month and year.
Calculate the number of steps required to solve the Collatz problem for
the month
These are my tasks that I need to fulfill, as Engineering student this how far I went in this. In the following codes I expect something like this
19
90
0
1
0
T M B B
The answer that I get is
Please enter your birthdate (dd mm yyyy): 12 11 1981
19
8468304
Press any key to continue . . .
8468304
How to get it right I know that my equation is right or(formula, method). However this is what I know.
#include<iostream>
#include<cstdlib>
using namespace std;
int main()
{
cout << "Please enter your birthdate (dd mm yyyy): ";
int day, month, year, count,rev;
int sum = 0;
cin >> day>> month >>year;
while (year!=0)
{
int count = year%10;
sum +=count;
year /= 10;
}
while(year>0)
{
rev = year%10;
year=year/10;
}
cout<<sum<<endl;
cout << rev;
system ("pause");
return 0;
}//end main
Please help!
After your first loop, while (year != 0), you don't reset the value of year, so it remains at zero and the second loop doesn't execute at all.
You need to save the value of year and use it when you start the second loop.
Just a note on organisation: I'd suggest to write a subroutine/function for every task, like
int digit_sum(int year) {
/* ... */
return sum;
}
int reverse_difference(int year) {
/* ... */
return diff;
}
and so on. This way you'll also prevent errors like modifying the year variable during the first calculation without saving the original value (which you did, as David Winant already pointed out).