For example this code is broken (I've just fixed it in actual code..).
uint64_t a = 1 << 60
It can be fixed as,
uint64_t a = (uint64_t)1 << 60
but then this passed my brain.
uint64_t a = UINT64_C(1) << 60
I know that UINT64_C(1) is a macro that expands usually as 1ul in 64-bit systems, but then what makes it different than just doing a type cast?
(uint64_t)1 is formally an int value 1 casted to uint64_t, whereas 1ul is a constant 1 of type unsigned long which is probably the same as uint64_t on a 64-bit system. As you are dealing with constants, all calculations will be done by the compiler and the result is the same.
The macro is a portable way to specify the correct suffix for a constant (literal) of type uint64_t. The suffix appended by the macro (ul, system specific) can be used for literal constants only.
The cast (uint64_t) can be used for both constant and variable values. With a constant, it will have the same effect as the suffix or suffix-adding macro, whereas with a variable of a different type it may perform a truncation or extension of the value (e.g., fill the higher bits with 0 when changing from 32 bits to 64 bits).
Whether to use UINT64_C(1) or (uint64_t)1 is a matter of taste. The macro makes it a bit more clear that you are dealing with a constant.
As mentioned in a comment, 1ul is a uint32_t, not a uint64_t on windows system. I expect that the macro UINT64_C will append the platform-specific suffix corresponding to uint64_t, so it might append uLL in this case. See also https://stackoverflow.com/a/52490273/10622916.