a/b does integer division. If either a or b is negative, the result depends on the compiler (rounding can go toward zero or toward negative infinity in pre-C99; in C99+, the rounding goes toward 0). The result has type int. floor(a/b) does the same division, converts the result to double, discards the (nonexistent) fractional part, and returns the result as a double.
a/b does integer division. If either a or b is negative, the result depends on the compiler (rounding can go toward zero or toward negative infinity in pre-C99; in C99+, the rounding goes toward 0). The result has type int. floor(a/b) does the same division, converts the result to double, discards the (nonexistent) fractional part, and returns the result as a double.
floor returns a double while a / b where both a and b are integers yields an integer value.
With the correct cast the value is the same.
If typeof operator existed in C (it does not) we would have:
(typeof (a /b)) floor(a / b) == a / b
EDIT: Now if the question is: is there any difference between:
(double) (a / b)
and
floor(a / (double) b)
the answer is yes. The results differ with respect to negative values.
Videos
You can expand the operator to be more than a single character. For example you could do /int~ and /float~ to disambiguate which division operation to use. It's pretty verbose but if you can find a shorter term to signify which operation you can use that.
This can then be expanded to other operations to for example ensure that overflow behavior is specified for integer addition +sat32~.
Some options that don't involve namespaces:
- Python:
a // b - Visual Basic:
a \ b - Pascal:
a div b
Normal division is a / b in all the cases.
When programming in C, how do I make the program to round correctly when only using integers for mathematic operation?
Example:
int x = 6;
int y = 8;
int division = x/y;
int myAnswer = 5 + division;
The accurate answer should be 5.75 so it should round up to 6. However, when you run this code, you'll get 5. How can you modify this code without using math.h or ceiling or floor function to round to the nearest value instead of 0?
Less assembly instructions in the generated code and quicker path to the result I think.
For the RISC machines with huge numbers of registers this one is better, as there are no branches at all and it is good for the pipeline and the cache.
For x86 actually it does not matter.
int floor_div3(int a, int b) {
int d = a / b;
return d * b == a ? d : d - ((a < 0) ^ (b < 0));
}
div() functions in standard C
I think you should look at the div() functions from <stdlib.h>. (They are standard C functions and are defined in all versions of the standard, despite the link to the POSIX specification.)
The C11 standard §7.22.6.2 specifies:
The
div… functions computenumer / denomandnumer % denomin a single operation.
Note that C11 specifies integer division in §6.5.5 (and C99 was similar):
When integers are divided, the result of the
/operator is the algebraic quotient with any fractional part discarded.105)105) This is often called "truncation toward zero".
but C90 (§6.3.5) was more flexible yet less useful:
When integers are divided and the division is inexact. if both operands are positive the result of the
/operator is the largest integer less than the algebraic quotient and the result of the%operator is positive. If either operand is negative, whether the result of the/operator is the largest integer less than or equal to the algebraic quotient or the smallest integer greater than or equal to the algebraic quotient is implementation-defined, as is the sign of the result of the%operator.
floor_div()
The computational code for the requested floor_div() using div() is neat and tidy.
int floor_div(int a, int b)
{
assert(b != 0);
div_t r = div(a, b);
if (r.rem != 0 && ((a < 0) ^ (b < 0)))
r.quot--;
return r.quot;
}
Test code
The print formatting in the code below is tailored rather precisely to the sample data. (It would be better, but more expansive, to use %4d and %-4d throughout). This code prints lines of length 89 characters plus newline; the more general layout would print lines of length 109. Neither avoids the horizontal scroll bar on SO.
#include <assert.h>
#include <stdio.h>
#include <stdlib.h>
static int floor_div(int a, int b)
{
assert(b != 0);
div_t r = div(a, b);
if (r.rem != 0 && ((a < 0) ^ (b < 0)))
r.quot--;
return r.quot;
}
static void test_floor_div(int n, int d)
{
assert(d != 0);
printf( "%3d/%-2d = %-3d (%3d)", +n, +d, floor_div(+n, +d), +n / +d);
printf("; %3d/%-3d = %-4d (%4d)", +n, -d, floor_div(+n, -d), +n / -d);
if (n != 0)
{
printf("; %4d/%-2d = %-4d (%4d)", -n, +d, floor_div(-n, +d), -n / +d);
printf("; %4d/%-3d = %-3d (%3d)", -n, -d, floor_div(-n, -d), -n / -d);
}
putchar('\n');
}
int main(void)
{
int numerators[] = { 0, 1, 2, 4, 9, 23, 291 };
enum { NUM_NUMERATORS = sizeof(numerators) / sizeof(numerators[0]) };
int denominators[] = { 1, 2, 3, 6, 17, 23 };
enum { NUM_DENOMINATORS = sizeof(denominators) / sizeof(denominators[0]) };
for (int i = 0; i < NUM_NUMERATORS; i++)
{
for (int j = 0; j < NUM_DENOMINATORS; j++)
test_floor_div(numerators[i], denominators[j]);
putchar('\n');
}
return 0;
}
Test output
0/1 = 0 ( 0); 0/-1 = 0 ( 0)
0/2 = 0 ( 0); 0/-2 = 0 ( 0)
0/3 = 0 ( 0); 0/-3 = 0 ( 0)
0/6 = 0 ( 0); 0/-6 = 0 ( 0)
0/17 = 0 ( 0); 0/-17 = 0 ( 0)
0/23 = 0 ( 0); 0/-23 = 0 ( 0)
1/1 = 1 ( 1); 1/-1 = -1 ( -1); -1/1 = -1 ( -1); -1/-1 = 1 ( 1)
1/2 = 0 ( 0); 1/-2 = -1 ( 0); -1/2 = -1 ( 0); -1/-2 = 0 ( 0)
1/3 = 0 ( 0); 1/-3 = -1 ( 0); -1/3 = -1 ( 0); -1/-3 = 0 ( 0)
1/6 = 0 ( 0); 1/-6 = -1 ( 0); -1/6 = -1 ( 0); -1/-6 = 0 ( 0)
1/17 = 0 ( 0); 1/-17 = -1 ( 0); -1/17 = -1 ( 0); -1/-17 = 0 ( 0)
1/23 = 0 ( 0); 1/-23 = -1 ( 0); -1/23 = -1 ( 0); -1/-23 = 0 ( 0)
2/1 = 2 ( 2); 2/-1 = -2 ( -2); -2/1 = -2 ( -2); -2/-1 = 2 ( 2)
2/2 = 1 ( 1); 2/-2 = -1 ( -1); -2/2 = -1 ( -1); -2/-2 = 1 ( 1)
2/3 = 0 ( 0); 2/-3 = -1 ( 0); -2/3 = -1 ( 0); -2/-3 = 0 ( 0)
2/6 = 0 ( 0); 2/-6 = -1 ( 0); -2/6 = -1 ( 0); -2/-6 = 0 ( 0)
2/17 = 0 ( 0); 2/-17 = -1 ( 0); -2/17 = -1 ( 0); -2/-17 = 0 ( 0)
2/23 = 0 ( 0); 2/-23 = -1 ( 0); -2/23 = -1 ( 0); -2/-23 = 0 ( 0)
4/1 = 4 ( 4); 4/-1 = -4 ( -4); -4/1 = -4 ( -4); -4/-1 = 4 ( 4)
4/2 = 2 ( 2); 4/-2 = -2 ( -2); -4/2 = -2 ( -2); -4/-2 = 2 ( 2)
4/3 = 1 ( 1); 4/-3 = -2 ( -1); -4/3 = -2 ( -1); -4/-3 = 1 ( 1)
4/6 = 0 ( 0); 4/-6 = -1 ( 0); -4/6 = -1 ( 0); -4/-6 = 0 ( 0)
4/17 = 0 ( 0); 4/-17 = -1 ( 0); -4/17 = -1 ( 0); -4/-17 = 0 ( 0)
4/23 = 0 ( 0); 4/-23 = -1 ( 0); -4/23 = -1 ( 0); -4/-23 = 0 ( 0)
9/1 = 9 ( 9); 9/-1 = -9 ( -9); -9/1 = -9 ( -9); -9/-1 = 9 ( 9)
9/2 = 4 ( 4); 9/-2 = -5 ( -4); -9/2 = -5 ( -4); -9/-2 = 4 ( 4)
9/3 = 3 ( 3); 9/-3 = -3 ( -3); -9/3 = -3 ( -3); -9/-3 = 3 ( 3)
9/6 = 1 ( 1); 9/-6 = -2 ( -1); -9/6 = -2 ( -1); -9/-6 = 1 ( 1)
9/17 = 0 ( 0); 9/-17 = -1 ( 0); -9/17 = -1 ( 0); -9/-17 = 0 ( 0)
9/23 = 0 ( 0); 9/-23 = -1 ( 0); -9/23 = -1 ( 0); -9/-23 = 0 ( 0)
23/1 = 23 ( 23); 23/-1 = -23 ( -23); -23/1 = -23 ( -23); -23/-1 = 23 ( 23)
23/2 = 11 ( 11); 23/-2 = -12 ( -11); -23/2 = -12 ( -11); -23/-2 = 11 ( 11)
23/3 = 7 ( 7); 23/-3 = -8 ( -7); -23/3 = -8 ( -7); -23/-3 = 7 ( 7)
23/6 = 3 ( 3); 23/-6 = -4 ( -3); -23/6 = -4 ( -3); -23/-6 = 3 ( 3)
23/17 = 1 ( 1); 23/-17 = -2 ( -1); -23/17 = -2 ( -1); -23/-17 = 1 ( 1)
23/23 = 1 ( 1); 23/-23 = -1 ( -1); -23/23 = -1 ( -1); -23/-23 = 1 ( 1)
291/1 = 291 (291); 291/-1 = -291 (-291); -291/1 = -291 (-291); -291/-1 = 291 (291)
291/2 = 145 (145); 291/-2 = -146 (-145); -291/2 = -146 (-145); -291/-2 = 145 (145)
291/3 = 97 ( 97); 291/-3 = -97 ( -97); -291/3 = -97 ( -97); -291/-3 = 97 ( 97)
291/6 = 48 ( 48); 291/-6 = -49 ( -48); -291/6 = -49 ( -48); -291/-6 = 48 ( 48)
291/17 = 17 ( 17); 291/-17 = -18 ( -17); -291/17 = -18 ( -17); -291/-17 = 17 ( 17)
291/23 = 12 ( 12); 291/-23 = -13 ( -12); -291/23 = -13 ( -12); -291/-23 = 12 ( 12)