Size of Character ('A') in C/C++

Size of character ('a') in C/C++

In C, the type of a character constant like 'a' is actually an int, with size of 4 (or some other implementation-dependent value). In C++, the type is char, with size of 1. This is one of many small differences between the two languages.

Difference between sizeof(char) and sizeof(char *)

char is a character and sizeof(char) is defined to be 1. (N1570 6.5.3.4 The sizeof and _Alignof operators, paragraph 4)

char* is a pointer to a character and sizeof(char*) depends on the environment. It is typically 4 in 32-bit environment and 8 in 64-bit environment.

In typical environment where sizeof(char*) > sizeof(char), malloc(sizeof(char*)*len + 1) will (at least try to) allocate more memory than malloc(sizeof(char)*len + 1) if len is small enough not to cause integer overflow.

Why is sizeof a char literal not the same as sizeof(char)? [duplicate]

In C opposite to C++ integer character constants (literals) have the type int.

So the value of the expression sizeof( 'a' ) is equal to the value of the expression sizeof( int ). While sizeof( char ) is always equal to 1.

From the C Standard (6.5.3.4 The sizeof and alignof operators)

4 When sizeof is applied to an operand that has type char, unsigned
char, or signed char, (or a qualified version thereof) the result is
1...

and (6.4.4.4 Character constants)

10 An integer character constant has type int. The value of an
integer character constant containing a single character that maps to
a single-byte execution character is the numerical value of the
representation of the mapped character interpreted as an integer. The
value of an integer character constant containing more than one
character (e.g., 'ab'), or containing a character or escape sequence
that does not map to a single-byte execution character, is
implementation-defined. If an integer character constant contains a
single character or escape sequence, its value is the one that results
when an object with type char whose value is that of the single
character or escape sequence is converted to type int.

Pay attention to that usually objects of the type char used in operations as operands or in expressions are converted to the type int due to the integer promotions.

Why is the size of a character literal in C different than in C++ [duplicate]

In C, 'i' has type int for backwards-compatibility reasons. Thus sizeof('i') shows the size of an int on the chosen compilation platform.

In C++, because overloading made it more urgent to avoid giving surprising types to expression, it was decided to break backwards compatibility and to give 'i' the type char.

In C, why is sizeof(char) 1, when 'a' is an int?

In C 'a' is an integer constant (!?!), so 4 is correct for your architecture. It is implicitly converted to char for the assignment. sizeof(char) is always 1 by definition. The standard doesn't say what units 1 is, but it is often bytes.

Single, double quotes and sizeof('a') in C/C++

In C, character constants such as 'a' have type int, in C++ it's char.

Regarding the last question, yes,

char ch = 'a';

causes an implicit conversion of the int to char.

sizeof('a') vs size('aa')

sizeof('a') which is type int

Neither one of 'a' or sizeof('a') is an int. In C++, one-character literals are of type char, and the type of a sizeof() expression is size_t.

Furthermore:

type int [...] should give me 4 bytes

No, int need not necessarily be exactly 4 bytes long.

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