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  lt  lt    m      lt  lt  m  lt  lt    n       return m

User · Answer

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

User · Answer

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

User · Answer

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

User · Answer

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 lt float gt  a y - b y     a x - b x

User · Answer

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

User · Answer

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

User · Answer

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

User · Answer

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

[c++] C++ int float casting

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