this code is to determine the number of dyas between two different dates and i hope i did it well. this ny solution

Works well, but the code organization and structure is very POOR!

I would recommend to start spacing better.
if you have a ; dont put more code behind it, go to another line, it gets very confusing to read.

Agreed. It makes it easier to read and understand for other's when it's organized. Put space in between functions, declarations etc..

Also try commenting where areas of the program may be complex or hard to understand.

Otherwise it looks good, Keep it up.

seems like it should work just fine. just as a side note:

the difftime(...) function will return the difference of two times in seconds, and knowing that there are 86400 seconds in a day, the same problem could be solved in this manner as well. This implimentation will only work if you are using two time_t objects.

#include<iostream>
#include<cmath>
using namespace std;
int mon[12]={31,28,31,30,31,30,31,31,30,31,30,31};

void GateDate ();

void CalDays();

int yr1,yr2,m1,m2,d1,mons2,mons1,d2,sum=0,yrs,mons,days,j;

int const loop = 365;
int main ()
{
GateDate();
CalDays();

system ("pause");
return 0;
}
void GateDate()

{
cout<<"Enter date in the form DAY - MONTH -YEAR NOTE DATE TWO > DATE ONE \n";
cout<<" Enter date ONE and press enter to enter month... \n";

cin>>d1>>m1>>yr1;

cout<<" Enter date TWO and press enter to enter month... \n";

cin>>d2>>m2>>yr2;
}


void CalDays()
{
days = d2 - d1;
yrs = yr2 - yr1;
if (m2>=m1)
{
mons = m1 - m2;
mons1 = mons;
mons2 = abs(mons);
}

else

{
mons = m2 - m1;
mons1 = mons;
mons2 = abs(mons);
}

sum = yrs * loop + mons + days + mons2;

if (m2>m1)
{
j = m1;
}

else
{
j = m2;
}

if ( mons2 != 0 )
{
for(int i = 0; i < mons2;i++)
{
sum = sum +mon[j];
j++;
}
}
cout<<"the number of days in \n"<<d1<<" / "<<m1<<" / "<<yr1<<"\n"<<d2<<" / "<<m2<<" / "<<yr2<<"\n\nis "<<(sum + mons )<<"\n";
}

You mention it does not work for some dates...can you give an example?