Implicit type conversion rules in C operators

Question

I want to be better about knowing when I should cast  What are the implicit type conversion rules in C   when adding  multiplying  etc  For example   int   float     int   float     float   int     int   float     float   int     int   int     int   float       et cetera     Will the expression always be evaluated as the more precise type  Do the rules differ for Java   Please correct me if I have worded this question inaccurately

User · Answer

If you exclude the unsigned types  there is an ordered hierarchy  signed char  short  int  long  long long  float  double  long double   First  anything coming before int in the above will be converted to int   Then  in a binary operation  the lower ranked type will be converted to the higher  and the results will be the type of the higher    You ll note that  from the hierarchy  anytime a floating point and an integral type are involved  the integral type will be converted to the floating point type    Unsigned complicates things a bit  it perturbs the ranking  and parts of the ranking become implementation defined   Because of this  it s best to not mix signed and unsigned in the same expression    Most C   experts seem to avoid unsigned unless bitwise operations are involved   That is  at least  what Stroustrup recommends

User · Answer

In C   operators  for POD types  always act on objects of the same type  Thus if they are not the same one will be promoted to match the other  The type of the result of the operation is the same as operands  after conversion     If either is      long          double the other is promoted to      long          double If either is                    double the other is promoted to                    double If either is                    float  the other is promoted to                    float If either is long long unsigned int    the other is promoted to long long unsigned int If either is long long          int    the other is promoted to long long          int If either is long      unsigned int    the other is promoted to long      unsigned int If either is long               int    the other is promoted to long               int If either is           unsigned int    the other is promoted to           unsigned int If either is                    int    the other is promoted to                    int Both operands are promoted to int   Note  The minimum size of operations is int  So short char are promoted to int before the operation is done   In all your expressions the int is promoted to a float before the operation is performed  The result of the operation is a float   int   float   gt   float   float   float int   float   gt   float   float   float float   int   gt   float   float   float int   float   gt   float   float   float float   int   gt   float   float   float int   int                       int int   float   gt    lt compiler error gt

User · Answer

Since the other answers don t talk about the rules in C  11 here s one  From C  11 standard  draft n3337    5 9  emphasized the difference       This pattern is called the usual arithmetic conversions  which are defined as follows           If either operand is of scoped enumeration type  no conversions are performed  if the other operand does not have the same type  the expression is ill-formed           If either operand is of type long double  the other shall be converted to long double           Otherwise  if either operand is double  the other shall be converted to double           Otherwise  if either operand is float  the other shall be converted to float           Otherwise  the integral promotions shall be performed on both operands  Then the following rules shall be applied to the promoted operands                If both operands have the same type  no further conversion is needed               Otherwise  if both operands have signed integer types or both have unsigned integer types  the     operand with the type of lesser integer conversion rank shall be converted to the type of the     operand with greater rank               Otherwise  if the operand that has unsigned integer type has rank greater than or equal to the     rank of the type of the other operand  the operand with signed integer type shall be converted to     the type of the operand with unsigned integer type               Otherwise  if the type of the operand with signed integer type can represent all of the values of the type of the operand with unsigned integer type  the operand with unsigned integer type shall     be converted to the type of the operand with signed integer type               Otherwise  both operands shall be converted to the unsigned integer type corresponding to the     type of the operand with signed integer type       See here for a list that s frequently updated

User · Answer

This answer is directed in large part at a comment made by  RafalDowgird       The minimum size of operations is int   - This would be very strange    what about architectures that efficiently support char short   operations   Is this really in the C   spec    Keep in mind that the C   standard has the all-important  as-if  rule  See section 1 8  Program Execution      3  This provision is sometimes called the  as-if  rule  because an   implementation  is free to disregard any requirement of the Standard   as long as the result is as if the requirement had been obeyed  as far   as can be determined from the observable behavior of the program    The compiler cannot set an int to be 8 bits in size  even if it were the fastest  since the standard mandates a 16-bit minimum int   Therefore  in the case of a theoretical computer with super-fast 8-bit operations  the implicit promotion to int for arithmetic could matter  However  for many operations  you cannot tell if the compiler actually did the operations in the precision of an int and then converted to a char to store in your variable  or if the operations were done in char all along   For example  consider unsigned char   unsigned char   unsigned char   unsigned char  where addition would overflow  let s assume a value of 200 for each   If you promoted to int  you would get 600  which would then be implicitly down cast into an unsigned char  which would wrap modulo 256  thus giving a final result of 88  If you did no such promotions you d have to wrap between the first two additions  which would reduce the problem from 200   200   200 to 144   200  which is 344  which reduces to 88  In other words  the program does not know the difference  so the compiler is free to ignore the mandate to perform intermediate operations in int if the operands have a lower ranking than int   This is true in general of addition  subtraction  and multiplication  It is not true in general for division or modulus

User · Answer

Whole chapter 4 talks about conversions  but I think you should be mostly interested in these    4 5 Integral promotions  conv prom  An rvalue of type char  signed char  unsigned char  short int  or unsigned short int can be converted to an rvalue of type int if int can represent all the values of the source type  other- wise  the source rvalue can be converted to an rvalue of type unsigned int  An rvalue of type wchar t  3 9 1  or an enumeration type  7 2  can be converted to an rvalue of the first of the following types that can represent all the values of its underlying type  int  unsigned int  long  or unsigned long  An rvalue for an integral bit-field  9 6  can be converted to an rvalue of type int if int can represent all the values of the bit-field  otherwise  it can be converted to unsigned int if unsigned int can rep- resent all the values of the bit-field  If the bit-field is larger yet  no integral promotion applies to it  If the bit-field has an enumerated type  it is treated as any other value of that type for promotion purposes  An rvalue of type bool can be converted to an rvalue of type int  with false becoming zero and true becoming one  These conversions are called integral promotions     4 6 Floating point promotion  conv fpprom  An rvalue of type float can be converted to an rvalue of type double  The value is unchanged  This conversion is called floating point promotion     Therefore  all conversions involving float - the result is float   Only the one involving both int - the result is int   int   int   int

User · Answer

Caveat   The conversions occur from left to right   Try this   int i   3  j   2  double k   33  cout  lt  lt  k   j   i  lt  lt  endl     prints 22 cout  lt  lt  j   i   k  lt  lt  endl     prints 0

User · Answer

The type of the expression  when not both parts are of the same type  will be converted to the biggest of both  The problem here is to understand which one is bigger than the other  it does not have anything to do with size in bytes    In expressions in which a real number and an integer number are involved  the integer will be promoted to real number  For example  in int   float  the type of the expression is float   The other difference are related to the capability of the type  For example  an expression involving an int and a long int will result of type long int

User · Answer

Arithmetic operations involving float results in float  int   float   float int   float   float float   int   float int   float   float float   int   float int   int   int  For more detail answer  Look at what the section   5 9 from the C   Standard says  Many binary operators that expect operands of arithmetic or enumeration type cause conversions and yield result types in a similar way  The purpose is to yield a common type  which is also the type of the result  This pattern is called the usual arithmetic conversions  which are defined as follows      If either operand is of type long double  the other shall be converted to long double      Otherwise  if either operand is double  the other shall be converted to double      Otherwise  if either operand is float  the other shall be converted to float      Otherwise  the integral promotions  4 5  shall be performed on both operands 54      Then  if either operand is unsigned long the other shall be converted to unsigned long      Otherwise  if one operand is a long int and the other unsigned int  then if a long int can represent all the values of an unsigned int  the unsigned int shall be converted to a long int  otherwise both operands shall be converted to unsigned long int      Otherwise  if either operand is long  the other shall be converted to long      Otherwise  if either operand is unsigned  the other shall be converted to unsigned   Note  otherwise  the only remaining case is that both operands are int

User · Answer

My solution to the problem got WA wrong answer   then i changed one of int to long long int and it gave AC accept   Previously  I was trying to do long long int    int   int  and after I rectify it to long long int    long long int   int  Googling I came up with    1  Arithmetic Conversions  Conditions for Type Conversion    Conditions Met ---  Conversion   Either operand is of type long double  ---  Other operand is converted to type long double  Preceding condition not met and either operand is of type double  ---  Other operand is converted to type double  Preceding conditions not met and either operand is of type float  ---  Other operand is converted to type float  Preceding conditions not met  none of the operands are of floating types   ---  Integral promotions are performed on the operands as follows    If either operand is of type unsigned long  the other operand is converted to type unsigned long  If preceding condition not met  and if either operand is of type long and the other of type unsigned int  both operands are converted to type unsigned long  If the preceding two conditions are not met  and if either operand is of type long  t he other operand is converted to type long  If the preceding three conditions are not met  and if either operand is of type unsigned int  the other operand is converted to type unsigned int  If none of the preceding conditions are met  both operands are converted to type int     2   Integer conversion rules   Integer Promotions     Integer types smaller than int are promoted when an operation is performed on them  If all values of the original type can be represented as an int  the value of the smaller type is converted to an int  otherwise  it is converted to an unsigned int  Integer promotions are applied as part of the usual arithmetic conversions to certain argument expressions  operands of the unary    -  and   operators  and operands of the shift operators    Integer Conversion Rank    No two signed integer types shall have the same rank  even if they have the same representation  The rank of a signed integer type shall be greater than the rank of any signed integer type with less precision  The rank of long long int shall be greater than the rank of long int  which shall be greater than the rank of int  which shall be greater than the rank of short int  which shall be greater than the rank of signed char  The rank of any unsigned integer type shall equal the rank of the corresponding signed integer type  if any  The rank of any standard integer type shall be greater than the rank of any extended integer type with the same width  The rank of char shall equal the rank of signed char and unsigned char  The rank of any extended signed integer type relative to another extended signed integer type with the same precision is implementation-defined but still subject to the other rules for determining the integer conversion rank  For all integer types T1  T2  and T3  if T1 has greater rank than T2 and T2 has greater rank than T3  then T1 has greater rank than T3   Usual Arithmetic Conversions    If both operands have the same type  no further conversion is needed  If both operands are of the same integer type  signed or unsigned   the operand with the type of lesser integer conversion rank is converted to the type of the operand with greater rank  If the operand that has unsigned integer type has rank greater than or equal to the rank of the type of the other operand  the operand with signed integer type is converted to the type of the operand with unsigned integer type  If the type of the operand with signed integer type can represent all of the values of the type of the operand with unsigned integer type  the operand with unsigned integer type is converted to the type of the operand with signed integer type  Otherwise  both operands are converted to the unsigned integer type corresponding to the type of the operand with signed integer type  Specific operations can add to or modify the semantics of the usual arithmetic operations

[c++] Implicit type conversion rules in C++ operators

Examples related to c++

Examples related to casting

Examples related to implicit