C++ int float casting

Question

Why is m always = 0? The x and y members of someClass are integers.

float getSlope(someClass a, someClass b)
{           
    float m = (a.y - b.y) / (a.x - b.x);
    cout << " m = " << m << "\n";
    return m;
}

Durra

Please review the answers you marked the wrong answer. (float) is no C++ it is C style. – Sphalerite

this is a classic issue among programming learners; one good thing about javascript is it avoids issues with nonsense "integer division" which should throw errors, rather than output incorrect numbers. Javascript all numbers are floating point – Bond

Answer 1

Integer division occurs, then the result, which is an integer, is assigned as a float. If the result is less than 1 then it ends up as 0.

You'll want to cast the expressions to floats first before dividing, e.g.

float m = static_cast<float>(a.y - b.y) / static_cast<float>(a.x - b.x);

Lamebrain

..And watch out for division by zero! – Pacifistic

@Pacifistic - It's not that dangerous to divide by 0, when we're talking about floating point numbers. You could safely divide any float number by 0.0 or -0.0, will give you inf. But yes, if it's unexpected, it will cause problems. – Capitulate

@Kiril Kirov: Yeah it won't cause an exception/crash like integer division-by-zero does, but it leaves you with either +INF, -INF or NaN which will probably cause the OP further problems when he tries to use m. – Pacifistic

@Lamebrain isn't it enough to cast any one of the elements of the expression to float? e.g. (static_cast<float>(a.y)-b.y)/(a.x-b.x) – Verism

You should use a C++-style cast instead of a C-style cast in C++. – Concordance

@Verism cast is useful to quiet a warning about the loss of precision withint to float, which typically needs to form a rounded float for large int values. – Damnatory

@Lamebrain thanks for writing such a concise answer! – Bond

@S.S. Anne: I finally fixed my answer since it's stayed accepted all these years. Little over 10 years and a month... better late than never, right? (I think I wasn't very active at the time you commented so I may have missed it.) – Lamebrain

If you're not familiar with them, you might benefit more from learning templates and variadic types/functions. It would be no problem to make a function that uses either static_cast or perfect forwarding on its arguments before calling back to... whatever it is you were doing. – Hollander

Answer 2

You need to use cast. I see the other answers, and they will really work, but as the tag is C++ I'd suggest you to use static_cast:

float m = static_cast< float >( a.y - b.y ) / static_cast< float >( a.x - b.x );

Capitulate

I don't necessarily disagree, but counterpoints: (1) this is far less readable, and (2) less portable – Deary

@Deary - "less portable" in what way? – Capitulate

Namely, it couldn't be ported directly to C... again, just a counterpoint :) – Deary

@Deary - C is completely different language. I don't qualify this as "portability issue", but I get your point – Capitulate

+1 this should be the correct answer as the OP asked for C++. stroustrup.com/bs_faq2.html#static-cast – Britannic

Why is everybody casting both operands? I thought if either n or d were floats, n/d would promote the other and do floating division, divf if you will. Personally I would lean toward making them doubles for the operation even if the destination is float (a cost of float's resolution near zero is that it has low resolution where large integers are still contiguous. Try casting the top few ints to float and back.) – Hollander

Answer 3

you can cast both numerator and denominator by multiplying with (1.0) .

Anzac

Ironically this answer avoids the unnecessary clunky syntax everyone else used and uses higher precision appropriately. Underrated. – Hollander

Answer 4

You should be aware that in evaluating an expression containing integers, the temporary results from each stage of evaluation are also rounded to be integers. In your assignment to float m, the value is only converted to the real-number capable float type after the integer arithmetic. This means that, for example, 3 / 4 would already be a "0" value before becoming 0.0. You need to force the conversion to float to happen earlier. You can do this by using the syntax float(value) on any of a.y, b.y, a.x, b.x, a.y - b.y, or a.x - b.x: it doesn't matter when it's done as long as one of the terms is a float before the division happens, e.g.

float m = float(a.y - b.y) / (a.x - b.x); 
float m = (float(a.y) - b.y) / (a.x - b.x); 
...etc...

Muff

(float(a.y) - b.y) / (a.x - b.x) is strange as it does a float subtraction for .y, yet an int subtraction for .x. – Damnatory

Answer 5

Because (a.y - b.y) is probably less then (a.x - b.x) and in your code the casting is done after the divide operation so the result is an integer so 0.

You should cast to float before the / operation

Twice

Answer 6

You are performing calculations on integers and assigning its result to float. So compiler is implicitly converting your integer result into float

Reach

Answer 7

When doing integer division, the result will always be a integer unless one or more of the operands are a float. Just type cast one/both of the operands to a float and the compiler will do the conversion. Type casting is used when you want the arithmetic to perform as it should so the result will be the correct data type.

float m = static_cast<float>(a.y - b.y) / (a.x - b.x);

Heteropolar

try to reEdit your answer by highlighting the code – Subinfeudate

Answer 8

he does an integer divide, which means 3 / 4 = 0. cast one of the brackets to float

 (float)(a.y - b.y) / (a.x - b.x);

Chelyuskin

Answer 9

You can use ios manipulators fixed(). It will allow you to print floating point values.

Negotiate

Answer 10

if (a.y - b.y) is less than (a.x - b.x), m is always zero.

so cast it like this.

float m = ((float)(a.y - b.y)) / ((float)(a.x - b.x));

Corri