C bitwise-shift: right operand considered for implicit type-conversion?

Question

gcc 4.8.4 warns about 1u << 63ul​ (assuming 64 bit long and 32 bit int) and computes 0. Rightfully so (no promotion from 1u​ to 1ul before shifting)?

ISO/IEC 9899:201x, 6.3.1.8 (Usual arithmetic conversions): "Many operators that expect operands of arithmetic type cause conversions"; 6.5.7 (Bitwise shift operators): "The integer promotions are performed on each of the operands...".

But I am not unable to conclude. Which are those "many operators"? As I understand, "integer promotion" does not pertain to types wider than int (am I correct?), but the standard does not explicitly state that the right operand of a bitwise-shift is not taken into account for the implicit type conversion.

Answer 1

Each operation documents this separately. For example, n1548 §6.5.5 "Multiplicative operators" ¶3

The usual arithmetic conversions are performed on the operands.

This phrase is omitted from §6.5.7 "Bitwise shift operators". Instead it says:

The integer promotions are performed on each of the operands. The type of the result is the type of the promoted left operand. …

Since the section on bitwise shift operators says nothing about "usual arithmetic conversions", that conversion does not happen.