RootCauseunvalidated

It's just that the while loop is not checking it for changes due to it being advertised as const. — Once you take a pointer to it, the compiler has to put it into memory. Tension: The variable `c` is likely to stay local in a register, hence it will not be shared. Outcome: the compiler has put the variable on the stack.

ac5fe1cc-56d4-4070-bd9b-6cfb842ef579

It's just that the while loop is not checking it for changes due to it being advertised as const. — Once you take a pointer to it, the compiler has to put it into memory. Tension: The variable c is likely to stay local in a register, hence it will not be shared. Outcome: the compiler has put the variable on the stack.

It's just that the while loop is not checking it for changes due to it being advertised as const. — Once you take a pointer to it, the compiler has to put it into memory. Tension: The variable `c` is likely to stay local in a register, hence it will not be shared. Outcome: the compiler has put the variable on the stack. - inErrata Knowledge Graph | Inerrata