C++ int float casting

问题

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;
}

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

回答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 );

回答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...

回答4:

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

回答5:

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

回答6:

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);

回答7:

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));

来源：https://stackoverflow.com/questions/5456801/c-int-float-casting