问题
Is the following code undefined behavior according to GCC in C99 mode:
signed char c = CHAR_MAX; // assume CHAR_MAX < INT_MAX
c = c + 1;
printf("%d", c);
回答1:
signed char overflow does cause undefined behavior, but that is not what happens in the posted code.
With c = c + 1, the integer promotions are performed before the addition, so c is promoted to int in the expression on the right. Since 128 is less than INT_MAX, this addition occurs without incident. Note that char is typically narrower than int, but on rare systems char and int may be the same width. In either case a char is promoted to int in arithmetic expressions.
When the assignment to c is then made, if plain char is unsigned on the system in question, the result of the addition is less than UCHAR_MAX (which must be at least 255) and this value remains unchanged in the conversion and assignment to c.
If instead plain char is signed, the result of the addition is converted to a signed char value before assignment. Here, if the result of the addition can't be represented in a signed char the conversion "is implementation-defined, or an implementation-defined signal is raised," according to §6.3.1.3/3 of the Standard. SCHAR_MAX must be at least 127, and if this is the case then the behavior is implementation-defined for the values in the posted code when plain char is signed.
The behavior is not undefined for the code in question, but is implementation-defined.
回答2:
No, it has implementation-defined behavior, either storing an implementation-defined result or possibly raising a signal.
Firstly, the usual arithmetic conversions are applied to the operands. This converts the operands to type int and so the computation is performed in type int. The result value 128 is guaranteed to be representable in int, since INT_MAX is guaranteed to be at least 32767 (5.2.4.2.1 Sizes of integer types), so next a value 128 in type int must be converted to type char to be stored in c. If char is unsigned, CHAR_MAX is guaranteed to be at least 255; otherwise, if SCHAR_MAX takes its minimal value of 127:
6.3.1.3 Signed and unsigned integers
When a value with integer type is converted to another integer type, [if] the new type is signed and the value cannot be represented in it[,] either the result is implementation-defined or an implementation-defined signal is raised.
In particular, gcc can be configured to treat char as either signed or unsigned (-f\[un\]signed-char); by default it will pick the appropriate configuration for the target platform ABI, if any. If a signed char is selected, all current gcc target platforms that I am aware of have an 8-bit byte (some obsolete targets such as AT&T DSP1600 had a 16-bit byte), so it will have range [-128, 127] (8-bit, two's complement) and gcc will apply modulo arithmetic yielding -128 as the result:
The result of, or the signal raised by, converting an integer to a signed integer type when the value cannot be represented in an object of that type (C90 6.2.1.2, C99 and C11 6.3.1.3).For conversion to a type of width N, the value is reduced modulo 2^N to be within range of the type; no signal is raised.
来源：https://stackoverflow.com/questions/47263865/is-signed-char-overflow-undefined-within-the-range-255-to-255