Question

After invoking longjmp(), non-volatile-qualified local objects should not be accessed if their values could have changed since the invocation of setjmp(). Their value in this case is considered indeterminate, and accessing them is undefined behavior.

Now my question is why volatile works in this situation? Wouldn’t change in that volatile variable still fail the longjmp? For example, how longjmp will work correctly in the example given below? When the code get backs to setjmp after longjmp, wouldn’t the value of local_var be 2 instead of 1?

void some_function()
{
  volatile int local_var = 1;

  setjmp( buf );
  local_var = 2;
  longjmp( buf, 1 );
}

Answer 1

setjmp and longjmp clobber registers. If a variable is stored in a register, its value gets lost after a longjmp.

Conversely, if it’s declared as volatile, then every time it gets written to, it gets stored back to memory, and every time it gets read from, it gets read back from memory every time. This hurts performance, because the compiler has to do more memory accesses instead of using a register, but it makes the variable’s usage safe in the face of longjmping.