How do promotion rules work when the signedness on either side of a binary operator differ? [duplicate]

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This is outlined explicitly in §5/9:

Many binary operators that expect operands of arithmetic or enumeration type cause conversions and yield result types in a similar way. The purpose is to yield a common type, which is also the type of the result. This pattern is called the usual arithmetic conversions, which are defined as follows:

  • If either operand is of type long double, the other shall be converted to long double.
  • Otherwise, if either operand is double, the other shall be converted to double.
  • Otherwise, if either operand is float, the other shall be converted to float.
  • Otherwise, the integral promotions shall be performed on both operands.
  • Then, if either operand is unsigned long the other shall be converted to unsigned long.
  • Otherwise, if one operand is a long int and the other unsigned int, then if a long int can represent all the values of an unsigned int, the unsigned int shall be converted to a long int; otherwise both operands shall be converted to unsigned long int.
  • Otherwise, if either operand is long, the other shall be converted to long.
  • Otherwise, if either operand is unsigned, the other shall be converted to unsigned.

[Note: otherwise, the only remaining case is that both operands are int]

In both of your scenarios, the result of operator+ is unsigned. Consequently, the second scenario is effectively:

int result = static_cast<int>(us + static_cast<unsigned>(neg));

Because in this case the value of us + neg is not representable by int, the value of result is implementation-defined – §4.7/3:

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.

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