Printing a char with printf

Question

Are both these codes the same  char ch    a   printf   d   ch     Will it  print a garbage value   I am confused about this  printf   d     0       Will this print 0 or garbage value  Because when i do this   printf   d   sizeof   n        It prints 4  Why is sizeof   n   4 bytes  The same thing in C   prints 1 bytes  Why is that   So here s the main question  in c language is printf   d     0   supposed to print 0  and in C   printf   d     0   supposed to print garbage

User · Answer

In C char gets promoted to int in expressions  That pretty much explains every question  if you think about it   Source  The C Programming Language by Brian W Kernighan and Dennis M Ritchie  A must read if you want to learn C   Also see this stack overflow page  where people much more experienced then me can explain it much better then I ever can

User · Answer

include  lt stdio h gt   include  lt stdlib h gt   int func char a  char b  char c     demonstration that char on stack is promoted to int                                         note  this promotion is NOT integer promotion  but promotion during handling of the stack  don t confuse the two        const char  p    amp a    printf  a  d n            b  d n            c  d n    p  p - int sizeof int    p - int sizeof int    2       don t do this  might probably work on x86 with gcc  but again  don t do this      int main void      func 1  2  3        printf with  d treats its argument as int  argument must be int or smaller - gt  works because of conversion to int when on stack -- see demo above    printf   d   d   d n    long long  1  2  3      don t do this  Argument must be int or smaller type  like char    which is converted to int when on the stack -- see above          backslash followed by number is a oct VALUE   printf   d n     377                   prints -1   - gt  IF char is signed char  char literal has all bits set and is thus value -1                                                       - gt  char literal is then integer promoted to int   this promotion has nothing to do with the stack  don t confuse the two                                                 prints 255  - gt  IF char is unsigned char  char literal has all bits set and is thus value 255                                                       - gt  char literal is then integer promoted to int           backslash followed by x is a hex VALUE   printf   d n     xff                   prints -1   - gt  IF char is signed char  char literal has all bits set and is thus value -1                                                       - gt  char literal is then integer promoted to int                                             prints 255  - gt  IF char is unsigned char  char literal has all bits set and is thus value 255                                                       - gt  char literal is then integer promoted to int        printf   d n   255                     prints 255     printf   d n    char 255               prints -1   - gt  255 is cast to char where it is -1   printf   d n     n                     prints 10   - gt  Ascii newline has VALUE 10  The char 10 is integer promoted to int 10   printf   d n   sizeof   n              prints 4    - gt  Ascii newline is char  but integer promoted to int  And sizeof int  is 4  on many architectures    printf   d n   sizeof  char   n        prints 1    - gt  Switch off integer promotion via cast     return 0

User · Answer

Yes  it prints GARBAGE unless you are lucky   VERY IMPORTANT   The type of the printf sprintf fprintf argument MUST match the associated format type char   If the types don t match and it compiles  the results are very undefined   Many newer compilers know about printf and issue warnings if the types do not match  If you get these warnings  FIX them   If you want to convert types for arguments for variable functions  you must supply the cast  ie  explicit conversion  because the compiler can t figure out that a conversion needs to be performed  as it can with a function prototype with typed arguments    printf   d n    int  ch    In this example  printf is being TOLD that there is an  int  on the stack  The cast makes sure that whatever thing sizeof returns   some sort of long integer  usually   printf will get an int   printf   d    int  sizeof   n

User · Answer

d prints an integer  it will print the ascii representation of your character  What you need is  c   printf   c   ch     printf   d     0    prints the ascii representation of   0   which is 0  by escaping 0 you tell the compiler to use the ascii value 0   printf   d   sizeof   n     prints 4 because a character literal is an int  in C  and not a char

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In C  character constant expressions such as   n  or  a  have type int  thus sizeof   n     sizeof  int    whereas in C   they have type char     The statement printf   d     0    should simply print 0  the type of the expression   0  is int  and its value is 0     The statement printf   d   ch   should print the integer encoding for the value in ch  for ASCII   a     97

User · Answer

This is supposed to print the ASCII value of the character  as  d is the escape sequence for an integer  So the value given as argument of printf is taken as integer when printed   char ch    a   printf   d   ch     Same holds for printf   d     0     where the NULL character is interpreted as the 0 integer   Finally  sizeof   n   is 4 because in C  this notation for characters stands for the corresponding ASCII integer  So   n  is the same as 10 as an integer   It all depends on the interpretation you give to the bytes

[c++] Printing a char with printf

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