Printing hexadecimal characters in C

Question

I m trying to read in a line of characters  then print out the hexadecimal equivalent of the characters   For example  if I have a string that is  0xc0 0xc0 abc123   where the first 2 characters are c0 in hex and the remaining characters are abc123 in ASCII  then I should get   c0 c0 61 62 63 31 32 33   However  printf using  x gives me  ffffffc0 ffffffc0 61 62 63 31 32 33   How do I get the output I want without the  ffffff   And why is it that only c0  and 80  has the ffffff  but not the other characters

User · Answer

You are probably storing the value 0xc0 in a char variable  what is probably a signed type  and your value is negative  most significant bit set   Then  when printing  it is converted to int  and to keep the semantical equivalence  the compiler pads the extra bytes with 0xff  so the negative int will have the same numerical value of your negative char  To fix this  just cast to unsigned char when printing   printf   x    unsigned char variable

User · Answer

Indeed  there is type conversion to int  Also you can force type to char by using  hhx specifier   printf   hhX   a     In most cases you will want to set the minimum length as well to fill the second character with zeroes   printf   02hhX   a     ISO IEC 9899 201x says      7 The length modifiers and their meanings are    hh Specifies that a following d  i  o  u  x  or X conversion specifier applies to a   signed char or unsigned char argument  the argument will have   been promoted according to the integer promotions  but its value shall be   converted to signed char or unsigned char before printing   or that   a following

User · Answer

You can use hh to tell printf that the argument is an unsigned char  Use 0 to get zero padding and 2 to set the width to 2  x or X for lower uppercase hex characters  uint8 t a   0x0a  printf  quot  02hhX quot   a      Prints  quot 0A quot  printf  quot 0x 02hhx quot   a      Prints  quot 0x0a quot    Edit  If readers are concerned about 2501 s assertion that this is somehow not the  correct  format specifiers I suggest they read the printf link again  Specifically   Even though  c expects int argument  it is safe to pass a char because of the integer promotion that takes place when a variadic function is called  The correct conversion specifications for the fixed-width character types  int8 t  etc  are defined in the header  lt cinttypes gt  C    or  lt inttypes h gt   C   although PRIdMAX  PRIuMAX  etc is synonymous with  jd   ju  etc    As for his point about signed vs unsigned  in this case it does not matter since the values must always be positive and easily fit in a signed int  There is no signed hexideximal format specifier anyway  Edit 2    quot when-to-admit-you re-wrong quot  edition   If you read the actual C11 standard on page 311  329 of the PDF  you find   hh  Specifies that a following d  i  o  u  x  or X conversion specifier applies to a signed char or unsigned char argument  the argument will have been promoted according to the integer promotions  but its value shall be converted to signed char or unsigned char before printing   or that a following n conversion specifier applies to a pointer to a signed char argument

User · Answer

You are probably printing from a signed char array  Either print from an unsigned char array or mask the value with 0xff  e g  ar i   amp  0xFF  The c0 values are being sign extended because the high  sign  bit is set

User · Answer

Try something like this   int main         printf   x  x  x  x  x  x  x  x n           0xC0  0xC0  0x61  0x62  0x63  0x31  0x32  0x33       Which produces this       foo  c0 c0 61 62 63 31 32 33

User · Answer

You are seeing the ffffff because char is signed on your system  In C  vararg functions such as printf will promote all integers smaller than int to int  Since char is an integer  8-bit signed integer in your case   your chars are being promoted to int via sign-extension   Since c0 and 80 have a leading 1-bit  and are negative as an 8-bit integer   they are being sign-extended while the others in your sample don t   char    int c0 - gt  ffffffc0 80 - gt  ffffff80 61 - gt  00000061   Here s a solution   char ch   0xC0  printf   x   ch  amp  0xff     This will mask out the upper bits and keep only the lower 8 bits that you want

User · Answer

You can create an unsigned char   unsigned char c   0xc5    Printing it will give C5 and not ffffffc5   Only the chars bigger than 127 are printed with the ffffff because they are negative  char is signed    Or you can cast the char while printing   char c   0xc5   printf   x    unsigned char c

[c] Printing hexadecimal characters in C

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