search for: expressionappear

...I found that the undefined behavior on integer overflow, is specified in 5 (5): "If during the evaluation of an expression, the result is not mathematically defined or not in the range of representable values for its type, the behavior is undefined, unless such an expression is a constant expressionappears where an integral constant expression is required (5.19)" This seems to apply to mathematical expressions, but not to explicit/implicit type conversions. Hence, code like int index = a + 100; A[index] = ... may result on 64bit architectures in a sext(trunc(index)) expression, that can b...

>> 3) If the destination type is signed, the value is unchanged if it >> can be represented in the destination type (and bit-field >> width); otherwise, the value is implementation-defined. >> ---------------------------------------------------- >> >> 4.7.3 suggest to me, that the standard does not define a result for >> '(signed char)