size of uint8 uint16 and uint32

Question

The code base uses data types like uint8 1 byte wide unsigned integer   uint16 2 byte wide unsigned integer   uint32 4 byte wide unsigned integer  etc   question  uint8 is same as char    uint16 is same as int    uint32 is same as long    uint64 is same as double     for example   uint8 c 20        then the sizeof c  should be 20- Isnt it    I wrote a small code in visual studio as shown below    include  lt stdio h gt   include  lt string h gt  typedef unsigned int uint32  typedef unsigned int uint8  int main         double a   1320 134      uint32 b      uint8 c 20        b    unsigned int a      c 3    b      printf  value is  d   c 3        return 1      but the size of c in debug mode is 50  Why is it showing like that

User · Answer

It s quite unclear how you are computing the size   the size in debug mode       Use printf     printf  the size of c is  u n    unsigned int  sizeof c     Normally you d print a size t value  which is the type sizeof returns  with  zu  but if you re using a pre-C99 compiler like Visual Studio that won t work   You need to find the typedef statements in your code that define the custom names like uint8 and so on  those are not standard so nobody here can know how they re defined in your code   New C code should use  lt stdint h gt  which gives you uint8 t and so on

User · Answer

uint8  uint16  uint32  and uint64 are probably Microsoft-specific types   As of the 1999 standard  C supports standard typedefs with similar meanings  defined in  lt stdint h gt   uint8 t  uint16 t  uint32 t  and uint64 t  I ll assume that the Microsoft-specific types are defined similarly  Microsoft does support  lt stdint h gt   at least as of Visual Studio 2010  but older code may use uint8 et al   The predefined types char  short  int et al have sizes that vary from one C implementation to another  The C standard has certain minimum requirements  char is at least 8 bits  short and int are at least 16  long is at least 32  and each type in that list is at least as wide as the previous type   but permits some flexibility  For example  I ve seen systems where int is 16  32  or 64 bits   char is almost always exactly 8 bits  but it s permitted to be wider  And plain char may be either signed or unsigned   uint8 t is required to be an unsigned integer type that s exactly 8 bits wide  It s likely to be a typedef for unsigned char  though it might be a typedef for plain char if plain char happens to be unsigned  If there is no predefined 8-bit unsigned type  then uint8 t will not be defined at all   Similarly  each uintN t type is an unsigned type that s exactly N bits wide   In addition   lt stdint h gt  defines corresponding signed intN t types  as well as int fastN t and int leastN t types that are at least the specified width   The  u intN t types are guaranteed to have no padding bits  so the size of each is exactly N bits  The signed intN t types are required to use a 2 s-complement representation   Although uint32 t might be the same as unsigned int  for example  you shouldn t assume that   Use unsigned int when you need an unsigned integer type that s at least 16 bits wide  and that s the  natural  size for the current system  Use uint32 t when you need an unsigned integer type that s exactly 32 bits wide    And no  uint64 or uint64 t is not the same as double  double is a floating-point type

[c] size of uint8, uint16 and uint32?

Examples related to c

Examples related to visual-studio