Storage of data type char

Question

Somewhat oddly, C treats character constants as type int rather than type char. For example, on an ASCII system with a 32-bit int and an 8-bit char, the code

char grade = 'B';

represents 'B' as the numerical value 66 stored in a 32-bit unit, but grade winds up with 66 stored in an 8-bit unit. Please explain this lines.

This is how C works. A character constant has a type `int` by C standard. — Eugene Sh., Feb 09 '17 at 16:31
Not necessarily 8 bit uint. Could be other width, also could be singed, all implementation defined. — user3528438, Feb 09 '17 at 16:38
@user3528438; The OP said "8-bit unit", not "8-bit uint", and explicitly stated the *assumption* that `int` is 32 bits, `char` is 8 bits, and the system uses ASCII. — Keith Thompson, Feb 09 '17 at 16:40

score 1 · Answer 1 · edited Feb 09 '17 at 16:43

1

grade is of type char, and it's initialized with an expression of type int. That's perfectly ordinary and legal, and the int value is implicitly converted to type char. A value of any arithmetic type can be implicitly converted to any other arithmetic type.

Such implicit conversions are why, 99% of the time, the fact that character constants are of type int is not a problem; such constants are converted as appropriate depending on the context.

(Note that in C++ character constants are of type char. Remember that C and C++ are two different languages.)

edited Feb 09 '17 at 16:43

Petr Skocik

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answered Feb 09 '17 at 16:40

Keith Thompson

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@PSkocik: Thanks for the edit. I knew what I meant, it just didn't correspond to what my fingers typed for me. – Keith Thompson Feb 09 '17 at 21:16

Storage of data type char

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