On Tue, 26 May 2020 at 04:34, Thamilarasu Kandasamy (thamil) via Gcc-help <gcc-help@xxxxxxxxxxx> wrote: > > Hi > > I could deduce what is happening based on the output of the program. > > I am looking for an affirmative answer which explains because of this type promotion rule involving unsigned long and unsigned int, promotion of low rank type to high rank type doesn't happen, please quote that rule. > > As per my understanding, type promotion from unsigned int to unsigned long int should have happened. But it doesn't, Yes it does. But promoting (unsigned)0x123 to unsigned long will promote 0x00000123 to 0x0000000000000123, which you seem to think means it is not promoted. It is promoted, but the value doesn't change, and the high bits are all zero. > hence I am trying to know which type promotion rules apply here, is it documented anywhere ? In the C standard, of course. You might also find https://en.wikipedia.org/wiki/Sign_extension useful. GCC does exactly what it's supposed to. Maybe https://cppinsights.io/s/7375972e will help you understand which promotions happen at which points. ~(size - 1) is a 32-bit int with a negative value. When it gets promoted to 64-bits the value gets "sign extended" to produce a 64-bit unsigned value that has the same bit pattern as (signed long)-99. In hex (int)0xffffff9d gets promoted to (unsigned long)0xffffffffffffff9d. ~(usize - 1) is a 32-bit unsigned int (with a positive value, of course, because it's unsigned). It gets promoted from (unsigned)0xffffff9d to (unsigned long)0x000000000xffffff9d. When you AND that with the pointer value, obviously half the bits of the pointer will be lost. ~((unsigned long)usize - 1) is a 64-bit unsigned int, so no promotion happens. But in this case the value is already 64-bits before the ~ operation, so you get ~(0x0000000000000064 - 1) which is ~0x0000000000000063 which is 0xffffffffffffff9c. GCC is doing the right thing, you're just not understanding what the right thing is.
