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Printing a char with printf

Tags:

c++

c

printf

sizeof

People also ask

How do I print %s in printf?

Generally, printf() function is used to print the text along with the values. If you want to print % as a string or text, you will have to use '%%'.

How do I print char data?

The printf() function uses the format specifier %s to print char * . The standard does not specify how char is implemented as signed or unsigned.

How can I print a '%' character in a printf format string?

To print a percent-sign character, use %%. For non decimal floating-point numbers, signed value having the form [-]0xh. hhhhp[sign]ddd, where h is a single hexadecimal digit, hhhh is one or more hexadecimal digits, ddd is one or more decimal digits, and sign is + or -.


%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.


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.


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).


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.


#include <stdio.h>
#include <stdlib.h>

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 = &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 -> 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   -> IF char is signed char: char literal has all bits set and is thus value -1.
                                                     -> char literal is then integer promoted to int. (this promotion has nothing to do with the stack. don't confuse the two!!!) */
                                      /* prints 255  -> IF char is unsigned char: char literal has all bits set and is thus value 255.
                                                     -> char literal is then integer promoted to int */


  // backslash followed by x is a hex VALUE
  printf("%d\n", '\xff');             /* prints -1   -> IF char is signed char: char literal has all bits set and is thus value -1.
                                                     -> char literal is then integer promoted to int */
                                      /* prints 255  -> IF char is unsigned char: char literal has all bits set and is thus value 255.
                                                     -> char literal is then integer promoted to int */


  printf("%d\n", 255);                // prints 255


  printf("%d\n", (char)255);          // prints -1   -> 255 is cast to char where it is -1
  printf("%d\n", '\n');               // prints 10   -> Ascii newline has VALUE 10. The char 10 is integer promoted to int 10
  printf("%d\n", sizeof('\n'));       // prints 4    -> 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    -> Switch off integer promotion via cast!

  return 0;
}