Why does dividing two int not yield the right value when assigned to double

Question

How come that in the following snippet  int a   7  int b   3  double c   0  c   a   b    c ends up having the value 2  rather than 2 3333  as one would expect  If a and b are doubles  the answer does turn to 2 333  But surely because c  already is a double it should have worked with integers   So how come int int double doesn t work

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Here it is   a  Dividing two ints performs integer division always  So the result of a b in your case can only be an int   If you want to keep a and b as ints  yet divide them fully  you must cast at least one of them to double   double a b or a  double b or  double a  double b   b  c is a double  so it can accept an int value on assignement  the int is automatically converted to double and assigned to c   c  Remember that on assignement  the expression to the right of   is computed first  according to rule  a  above  and without regard of the variable to the left of    and then assigned to the variable to the left of    according to  b  above   I believe this completes the picture

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In C   language the result of the subexpresison is never affected by the surrounding context  with some rare exceptions   This is one of the principles that the language carefully follows  The  expression c   a   b contains of an independent subexpression a   b  which is interpreted independently from anything outside that subexpression  The language does not care that you later will assign the result to a double  a   b is an integer division  Anything else does not matter  You will see this principle followed in many corners of the language specification  That s juts how C    and C  works   One example of an exception I mentioned above is the function pointer assignment initialization in situations with function overloading  void foo int   void foo double    void   p  double     amp foo     automatically selects  foo fouble     This is one context where the left-hand side of an assignment initialization affects the behavior of the right-hand side   Also  reference-to-array initialization prevents array type decay  which is another example of similar behavior   In all other cases the right-hand side completely ignores the left-hand side

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This is technically a language-dependent  but almost all languages treat this subject the same   When there is a type mismatch between two data types in an expression  most languages will try to cast the data on one side of the   to match the data on the other side according to a set of predefined rules   When dividing two numbers of the same type  integers  doubles  etc   the result will always be of the same type  so  int int  will always result in int    In this case you have double var   integer result which casts the integer result to a double after the calculation in which case the fractional data is already lost   most languages will do this casting to prevent type inaccuracies without raising an exception or error    If you d like to keep the result as a double you re going to want to create a situation where you have double var   double result  The easiest way to do that is to force the expression on the right side of an equation to cast to double   c   a  double b  Division between an integer and a double will result in casting the integer to the double  note that when doing maths  the compiler will often  upcast  to the most specific data type this is to prevent data loss    After the upcast  a will wind up as a double and now you have division between two doubles   This will create the desired division and assignment   AGAIN  please note that this is language specific  and can even be compiler specific   however almost all languages  certainly all the ones I can think of off the top of my head  treat this example identically

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The   operator can be used for integer division or floating point division  You re giving it two integer operands  so it s doing integer division and then the result is being stored in a double

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The important thing is one of the elements of calculation be a float-double type  Then to get a double result you need to cast this element like shown below  c   static cast lt double gt  a    b   or c   a   static cast b   Or you can create it directly   c   7 0   3   Note that one of elements of calculation must have the   0  to indicate a division of a float-double type by an integer  Otherwise  despite the c variable be a double  the result will be zero too  an integer

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When you divide two integers  the result will be an integer  irrespective of the fact that you store it in a double

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For the same reasons above  you ll have to convert one of  a  or  b  to a double type  Another way of doing it is to use  double c    a 0 0  b   The numerator is  implicitly  converted to a double because we have added a double to it  namely 0 0

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With very few exceptions  I can only think of one   C   determines the entire meaning of an expression  or sub-expression  from the expression itself   What you do with the results of the expression doesn t matter  In your case  in the expression a   b  there s not a double in sight  everything is int   So the compiler uses integer division  Only once it has the result does it consider what to do with it  and convert it to double

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This is because you are using the integer division version of operator   which takes 2 ints and returns an int   In order to use the double version  which returns a double  at least one of the ints must be explicitly casted to a double   c   a  double b

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c is a double variable  but the value being assigned to it is an int value because it results from the division of two ints  which gives you  integer division   dropping the remainder   So what happens in the line c a b is   a b is evaluated  creating a temporary of type int the value of the temporary is assigned to c after conversion to type double    The value of a b is determined without reference to its context  assignment to double

[c++] Why does dividing two int not yield the right value when assigned to double?

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