请解释sizeof（）的输出

Question

I am new in c programming, and I am trying to understand the sizeof function.

Please help me understand how this program works:

#include<stdio.h>
main( )
{
    printf ( "\n%d %d %d", sizeof ( '3'), sizeof ( "3" ), sizeof ( 3 ) ) ;
}

I am getting output as 4 2 4 .

However, I am not able to understand the reason I get this output. Kindly explain it.

Answer 1

1. sizeof ( '3') , it is the size of character constant which is int so you are getting value as 4 on your machine.

2. sizeof ( "3" ) , it is size of string ie 2 .
   String "3" is made of 2 character ('3'+'\\0') = "3" . And we know sizeof(char) is 1.

3. sizeof ( 3 ) , it is size of int which is 4 on your machine.

Answer 2

First off, a couple of notes:

• You should not omit the return type of main. See What should main() return in C and C++?
• Enable warnings. On GCC, you would generally use -Wall . This should give you a warning about your format string.
• In C99, you should use %zu instead of %d as explained in How to print size_t variable portably? . The C standard says the type of the result of sizeof is size_t (§6.5.3.4/5). It is undefined behavior if your conversion specifiers don't match your arguments.

Onto your question.

• In the comp.lang.c FAQ , question 8.9 explains that character constants in C are of type int , so sizeof('3') is sizeof(int) (which appears to be 4 on your machine.)

• When applied to arrays, sizeof returns the size of the array in bytes. For example, sizeof(int[10]) returns 40 on a machine where sizeof(int) is 4. Since sizeof(char) is 1, it will equal the amount of characters in a string literal. As explained by Jonathan Leffler :

  1. sizeof("f") must return 2, one for the 'f' and one for the terminating '\\0'.

  2. [...]

  The string literal has the type 'array of size N of char' where N includes the terminal null.

  Remember, arrays do not decay ¹ to pointers when passed to sizeof.

  _{1 Exception to array not decaying into a pointer?}

• And lastly, sizeof(3) is sizeof(int) .