How to Make the Division of 2 Ints Produce a Float Instead of Another Int

How to make the division of 2 ints produce a float instead of another int?

Just cast one of the two operands to a float first.

v = (float)s / t;

The cast has higher precedence than the division, so happens before the division.

The other operand will be effectively automatically cast to a float by the compiler because the rules say that if either operand is of floating point type then the operation will be a floating point operation, even if the other operand is integral. Java Language Specification, §4.2.4 and §15.17

Why dividing two integers doesn't get a float?

This is because of implicit conversion. The variables b, c, d are of float type. But the / operator sees two integers it has to divide and hence returns an integer in the result which gets implicitly converted to a float by the addition of a decimal point. If you want float divisions, try making the two operands to the / floats. Like follows.

#include <stdio.h>

int main() {
    int a;
    float b, c, d;
    a = 750;
    b = a / 350.0f;
    c = 750;
    d = c / 350;
    printf("%.2f %.2f", b, d);
    // output: 2.14 2.14
    return 0;
}

Dividing two integers to produce a float result

Cast the operands to floats:

float ans = (float)a / (float)b;

Integer division: How do you produce a double?

double num = 5;

That avoids a cast. But you'll find that the cast conversions are well-defined. You don't have to guess, just check the JLS. int to double is a widening conversion. From §5.1.2:

Widening primitive conversions do not
lose information about the overall
magnitude of a numeric value.

[...]

Conversion of an int or a long value
to float, or of a long value to
double, may result in loss of
precision-that is, the result may lose
some of the least significant bits of
the value. In this case, the resulting
floating-point value will be a
correctly rounded version of the
integer value, using IEEE 754
round-to-nearest mode (§4.2.4).

5 can be expressed exactly as a double.

integer division to float result

When you divide two ints you perform integer division, which, in this case will result in 22/64 = 0. Only once this is done are you creating a float. And the float representation of 0 is 0.0. If you want to perform floating point division, you should cast before dividing:

ws = ((float) zahl1) / i;

Why does integer division yield a float instead of another integer?

Take a look at PEP-238: Changing the Division Operator

The // operator will be available to request floor division unambiguously.

How can I force division to be floating point? Division keeps rounding down to 0?

In Python 2, division of two ints produces an int. In Python 3, it produces a float. We can get the new behaviour by importing from __future__.

>>> from __future__ import division
>>> a = 4
>>> b = 6
>>> c = a / b
>>> c
0.66666666666666663

How to get a float result by dividing two integer values using T-SQL?

The suggestions from stb and xiowl are fine if you're looking for a constant. If you need to use existing fields or parameters which are integers, you can cast them to be floats first:

SELECT CAST(1 AS float) / CAST(3 AS float)

or

SELECT CAST(MyIntField1 AS float) / CAST(MyIntField2 AS float)

Why does float divided by an integer return a float

Because of C's Usual arithmetic conversions.

Quoting the ANSI C standard § 3.2.1.5 Usual arithmetic conversions:

Many binary operators that expect operands of arithmetic type cause conversions and yield result types in a similar way. The purpose is to yield a common type, which is also the type of the result. This pattern is called the usual arithmetic conversions: First, if either operand has type long double, the other operand is converted to long double . [...] Otherwise, if either operand has type float, the other operand is converted to float.

Emphasis mine.

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