How to print an unsigned char in C

Question

I am trying to print char as positive value   char ch   212  printf   u   ch     but I get   4294967252   How I can get 212 in the output

User · Answer

In case you cannot change the declaration for whatever reason  you can do   char ch   212  printf   d    unsigned char  ch

User · Answer

There are two bugs in this code  First  in most C implementations with signed char  there is a problem in char ch   212 because 212 does not fit in an 8-bit signed char  and the C standard does not fully define the behavior  it requires the implementation to define the behavior   It should instead be  unsigned char ch   212   Second  in printf  quot  u quot  ch   ch will be promoted to an int in normal C implementations  However  the  u specifier expects an unsigned int  and the C standard does not define behavior when the wrong type is passed  It should instead be  printf  quot  u quot    unsigned  ch

User · Answer

This is because in this case the char type is signed on your system   When this happens  the data gets sign-extended during the default conversions while passing the data to the function with variable number of arguments  Since 212 is greater than 0x80  it s treated as negative   u interprets the number as a large positive number   212   0xD4   When it is sign-extended  FFs are pre-pended to your number  so it becomes  0xFFFFFFD4   4294967252   which is the number that gets printed   Note that this behavior is specific to your implementation  According to C99 specification  all char types are promoted to  signed  int  because an int can represent all values of a char  signed or unsigned      6 1 1 2  If an int can represent all values of the original type  the value is converted to an int  otherwise  it is converted to an unsigned int    This results in passing an int to a format specifier  u  which expects an unsigned int   To avoid undefined behavior in your program  add explicit type casts as follows   unsigned char ch    unsigned char 212  printf   u    unsigned int ch        In general  the standard leaves the signedness of char up to the implementation  See this question for more details

User · Answer

Declare your ch as  unsigned char ch   212     And your printf will work

User · Answer

Because char is by default signed declared that means the range of the variable is      -127 to  127     your value is overflowed  To get the desired value you have to declared the unsigned modifier  the modifier s  unsigned  range is     0 to 255   to get the the range of any data type follow the process 2 bit example charis 8 bit length to get its range just 2   power  8

User · Answer

The range of char is 127 to -128  If you assign 212  ch stores -44  212-128-128  not 212 So if you try to print a negative number as unsigned you get  MAX value of unsigned int -abs number  which in this case is 4294967252  So if you want to store 212 as it is in ch the only thing you can do is declare ch as   unsigned char ch    now the range of ch is 0 to 255

[c] How to print an unsigned char in C?

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