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;
}

Solution 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);

Solution 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&lt float &gt( a.y - b.y ) / static_cast&lt float &gt( a.x - b.x );

Solution 3

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...

Comments

[django11]

..And watch out for division by zero!

[Kiril Kirov]

@GrahamS - 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.

[django11]

@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.

[juanchopanza]

@BoltClock 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)

[notlesh]

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

[Kiril Kirov]

@notlesh - "less portable" in what way?

[notlesh]

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

[Kiril Kirov]

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

[Anubis]

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

[S.S. Anne]

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

[chux - Reinstate Monica]

@juanchopanza 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.

[chux - Reinstate Monica]

(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.

[AAEM]

try to reEdit your answer by highlighting the code

[cmarangu]

@BoltClock thanks for writing such a concise answer!

[BoltClock]

@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.)

[Michael Rovinsky]

Please add some code to illustrate your answer

[0xB00B]

I have a 30000 characters long Equation in my code, I can't be manually casting each term, is there a way to cast all ints to float in a that Equation ? (except using some python script to fix it, I already did that, but I am hoping for a proper way ?)