Standard C++ Behavior of Signed Char to Unsigned Int Conversion

Question

I quickly scanned through the C++03 standard but still cannot tell if this behavior is guaranteed:

signed char cNegOne=-1; //char is 8bits
unsigned int a=cNegOne; //int is 32 bits in my Windows system
printf("0x%x\n",a);

result is:

0xffffffff

VC++ gives 0xffffffff in 32-bit Windows. But my assumption is that the conversion could happen in 2 ways:

1) 8-bit signed char -1 is first directly converted to a 8-bit unsigned value which is binary 11111111 or decimal 255 which then widened to 32-bit unsigned int giving 255 (0xff) too.

2) 8-bit signed char -1 is signed extended to 32 bit signed int giving 0xffffffff then reinterpreted as 32 bit unsigned int.

Obviously the second way is used here. But why is it the case? In the standard, I cannot find anything talking about this. Is it implementation specific?

EDIT: the original text from C++03 Chapter 4

Standard conversions are implicit conversions defined for built-in types. Clause 4 enumerates the full set of such conversions. A standard conversion sequence is a sequence of standard conversions in the following order:

— Zero or one conversion from the following set: lvalue-to-rvalue conversion, array-to-pointer conversion, and function-to-pointer conversion.

— Zero or one conversion from the following set: integral promotions, floating point promotion, integral conversions, floating point conversions, floating-integral conversions, pointer conversions, pointer to member conversions, and boolean conversions.

— Zero or one qualification conversion.

Pay attention that the guaranteed order is the l to rvalue conversion (etc) happens before the set integral promotions/conversions but that doesn't mean integral promtions must happen before conversions - they're just in the same set. Or is my interpretation correct?

You can't have a promotion and a conversion both happen inside the same standard conversion sequence ("zero or one"). char to unsigned int is one integral conversion. — T.C.
– T.C., Commented Dec 7, 2016 at 16:36

Sneftel · Accepted Answer · 2016-12-07 16:20:20Z

5

When performing integer arithmetic in C and C++, the very first step is integer promotion. signed chars and shorts become signed ints; unsigned chars and shorts become unsigned ints. That's true even if you're, say, adding two shorts together: The operation itself is performed on ints, not shorts.

So when you think about integer conversion, think in terms of promotion followed by signedness changing. In the C++ standard, these two steps are described as one. That is, a separate integral promotion doesn't technically occur, but the effect of changing signedness if the destination type is wider than the source type is identical to that which would occur if an explicit promotion preceded the signedness change. The basic rule is that the result must be the input modulo 2^n where n is the number of bits of the output type. For an input of -1, and a 32-bit output type, that means an output of 2^32 - 1.

edited Dec 7, 2016 at 16:20

answered Dec 7, 2016 at 7:35

Sneftel

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