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i just started taking cpp classes in school and i got a new assignment. in this assignment i have to calculate how much it will cost for a customers carpet(after all the input).I also must use one function in this program. so i was wondering if there was a way i could make a function round up to the next number (ie if n=1.11 round it to n=2 or something). plz include examples.
thanks :)
im still pretty new to C++ myself, and usually helping others is no place for someone with my knowlege level, but still....
why not just make it simple. such as:
//...
a = x / y;
/////////////
b = x % y;
if ( b > 0 )
a = ++a;
/////////////
printf("%d", a);
thats assuming that 'a' is an int, and when you divide x by y, it rounds down for a
i dont know if that helped even in the slightest, but i figured i would try to do my part.
Code:#include <stdio.h>
int main(void)
{
float b = 1.90;
printf("%.0f",b); // %.0f means no decimal places.
return 0;
}
thanks for the quick replies!
But i was wondering if i could mak eit round up all the time.
(1.10=2 and 1.7=2)
(int) rounds down all the time (truncates), so this should always round up:
float Var;
int AnotherVar=(int)Var+1;
no offense to the fine examples but .....
According to the MSDN help files... use them and you rely on this board much less...
/* FLOOR.C: This example displays the largest integers
* less than or equal to the floating-point values 2.8
* and -2.8. It then shows the smallest integers greater
* than or equal to 2.8 and -2.8.
*/
#include <math.h>
#include <stdio.h>
void main( void )
{
double y;
y = floor( 2.8 );
printf( "The floor of 2.8 is %f\n", y );
y = floor( -2.8 );
printf( "The floor of -2.8 is %f\n", y );
y = ceil( 2.8 );
printf( "The ceil of 2.8 is %f\n", y );
y = ceil( -2.8 );
printf( "The ceil of -2.8 is %f\n", y );
}
You might also want to check
if (floor(var) == var) before you go typecasting int on your float.
Take the following example
floar var;
cin >> var;
cout >> (int)var+1;
If the user enters an integer number, casting int on it and adding 1 won't give you a desired result, as it would already be an int.