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That's undefined behavior, the compiler can do whatever it pleases to. It can even print 666, and be done.

No it's not. I had the same thought, but the author carefully points out from the standard: "The usual arithmetic conversions ensure that two operands to a “+” operator of type signed char are both promoted to signed int before the addition is performed."

He does assume sizeof(int) > sizeof(char), which is true on all platforms he has tried. It would be undefined on an AVR or other microcontroller where sizeof(int) == sizeof(char) though.

Thanks! Should read the article more carefully.

undefined on an AVR or other microcontroller where sizeof(int) == sizeof(char) though.

Just noting that ints are 16 bits wide in AVR-GCC, unless you use the -mint8 option which violates C standards.

Absolutely true. Note, though, that sizeof(int) == sizeof(char) is allowed if you have 16-bit chars.

16-bit chars aren't allowed in modern versions of C -- CHAR_BIT is required to be 8.

Thanks for the update! (For the curious: C99 or POSIX both require CHAR_BIT == 8.)

C99 only requires CHAR_BIT to be at least 8 (5.2.4.2.1:1). POSIX requires it to be exactly 8.

Oops, quite right. I missed the "equal or greater in magnitude" line when I was reading that section.

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