C++ int 浮点数转换

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;
}

Answer 1

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 );

Answer 2

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, eg

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

Answer 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 ay , by , ax , bx , ay - by , or ax - bx : it doesn't matter when it's done as long as one of the terms is a float before the division happens, eg

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

Answer 4

Because (ay - by) is probably less then (ax - bx) 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

Answer 5

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

Answer 6

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);

Answer 7

您可以通过乘以 (1.0) 来转换分子和分母。

Answer 8

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

Answer 9

他做了一个整数除法，这意味着 3 / 4 = 0。将其中一个括号转换为浮动

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

Answer 10

if (ay - by) is less than (ax - bx), m is always zero.

so cast it like this.

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