Understanding the sizeof operator in C

Question

char c = 'A';
printf("%d\n",sizeof(c));// output = 1
printf("%d\n",sizeof('A')); // output = 4

Why the sizeof operator gives different output for same character? Please Help

Because 'A' doesn't have a type. It's being interpreted as an `int`. — David Hoelzer, Dec 23 '15 at 03:49
Fair enough. Since 'A' isn't cast in any way it is an `int` by definition. I was focused more on the fact that there was a difference in how they were defined. Thanks! — David Hoelzer, Dec 23 '15 at 04:14
There's no such thing as an expression without a type ... what a weird concept. — Jim Balter, Dec 23 '15 at 05:46

score 6 · Accepted Answer · answered Dec 23 '15 at 03:41

6

c is a variable of type char; its size is 1 byte.

'A' is an int literal - don't ask me why the standard says that. Its size is 4 bytes on your platform (the same as sizeof(1)).

answered Dec 23 '15 at 03:41

user253751

Back in the very olden days, people used multibyte characters, like 'xy'. The PDP11 UNIX code had a lot of that stuff. It mostly went away when they ported C to bigendian machines. C++ fixed this ... character constants are chars. This is one of the few ways in which C is not a proper subset of C++. – Jim Balter Dec 23 '15 at 05:49

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