uint8 t vs unsigned char

Question

What is the advantage of using uint8 t over unsigned char in C   I know that on almost every system uint8 t is just a typedef for unsigned char  so why use it

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The whole point is to write implementation-independent code  unsigned char is not guaranteed to be an 8-bit type  uint8 t is  if available

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On almost every system I ve met uint8 t    unsigned char  but this is not guaranteed by the C standard   If you are trying to write portable code and it matters exactly what size the memory is  use uint8 t   Otherwise use unsigned char

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It documents your intent - you will be storing small numbers  rather than a character   Also it looks nicer if you re using other typedefs such as uint16 t or int32 t

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There s little  From portability viewpoint  char cannot be smaller than 8 bits  and nothing can be smaller than char  so if a given C implementation has an unsigned 8-bit integer type  it s going to be char  Alternatively  it may not have one at all  at which point any typedef tricks are moot   It could be used to better document your code in a sense that it s clear that you require 8-bit bytes there and nothing else  But in practice it s a reasonable expectation virtually anywhere already  there are DSP platforms on which it s not true  but chances of your code running there is slim  and you could just as well error out using a static assert at the top of your program on such a platform

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In my experience there are two places where we want to use uint8 t to mean 8 bits  and uint16 t  etc  and where we can have fields smaller than 8 bits  Both places are where space matters and we often need to look at a raw dump of the data when debugging and need to be able to quickly determine what it represents   The first is in RF protocols  especially in narrow-band systems  In this environment we may need to pack as much information as we can into a single message  The second is in flash storage where we may have very limited space  such as in embedded systems   In both cases we can use a packed data structure in which the compiler will take care of the packing and unpacking for us    pragma pack 1  typedef struct     uint8 t    flag1 1    uint8 t    flag2 1    padding1   reserved 6      not necessary but makes this struct more readable      uint32 t   sequence no    uint8 t    data 8     uint32 t   crc32    s mypacket   attribute    packed     pragma pack     Which method you use depends on your compiler  You may also need to support several different compilers with the same header files  This happens in embedded systems where devices and servers can be completely different - for example you may have an ARM device that communicates with an x86 Linux server   There are a few caveats with using packed structures  The biggest gotcha is that you must avoid dereferencing the address of a member  On systems with mutibyte aligned words  this can result in a misaligned exception - and a coredump   Some folks will also worry about performance and argue that using these packed structures will slow down your system  It is true that  behind the scenes  the compiler adds code to access the unaligned data members  You can see that by looking at the assembly code in your IDE   But since packed structures are most useful for communication and data storage then the data can be extracted into a non-packed representation when working with it in memory  Normally we do not need to be working with the entire data packet in memory anyway   Here is some relevant discussion   pragma pack 1  nor   attribute     aligned  1    works  Is gcc  39 s   attribute    packed      pragma pack unsafe   http   solidsmoke blogspot ca 2010 07 woes-of-structure-packing-pragma-pack html

User · Answer

Just to be pedantic  some systems may not have an 8 bit type  According to Wikipedia      An implementation is required to define exact-width integer types for N   8  16  32  or 64 if and only if it has any type that meets the requirements  It is not required to define them for any other N  even if it supports the appropriate types    So uint8 t isn t guaranteed to exist  though it will for all platforms where 8 bits   1 byte  Some embedded platforms may be different  but that s getting very rare  Some systems may define char types to be 16 bits  in which case there probably won t be an 8-bit type of any kind   Other than that  minor  issue   Mark Ransom s answer is the best in my opinion  Use the one that most clearly shows what you re using the data for   Also  I m assuming you meant uint8 t  the standard typedef from C99 provided in the stdint h header  rather than uint 8  not part of any standard

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That is really important for example when you are writing a network analyzer   packet headers are defined by the protocol specification  not by the way a particular platform s C compiler works

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As you said   almost every system    char is probably one of the less likely to change  but once you start using uint16 t and friends  using uint8 t blends better  and may even be part of a coding standard

[c] uint8_t vs unsigned char

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