How Do I Convert This To Inequality Notation -3 4]

What is an intuitive explanation of log likelihood ratio?

The log likelihood It is little risky  to answer intuitively the question asked. I like to use my rigorous Statistics training = 10yrs., practicing what I learned, in teaching, researching and applying for x yrs. and consulting for about y years with and without fees. Now in my retirement, I feel it is safe to keep on practicing it any way.  Having documented that, I will try to answer what you asked. The Maximum Likelihood method is used in estimating a parameter  θ of a distribution with a density function ∱(x), from samples drawn at random from it. Naturally, one wants the estimate to be the best possible one of the unknown  θ. So, a Likelihood function is defined such that it allows the sample to arrive at that desirable estimate. And we want that Likelihood to be Maximum for the best estimate. So, by definition the Likelihood Function is given by, L(θ)=∏∱(xi),i=1,2....n for a sample of size n. If one has a density function from the Type III family, such as, the Normal  or Gamma,  Weibull etc. the actual computation of the likelihood becomes rather long and complicated. So, we fall back on the natural log function and try to simplify the L(θ)=∏∱(xi),i=1,2....n, in general. That means, the notation for the continuos product is changed to Summation. or ∏→Σ And that is easier to handle. [ note: if you don't agree, try with the normal density, you will definitely like it]Then, we have the expression of log likelihood which needs to be maximized. So from the Freshman level Calculus, one maximizes by equating its first derivative ( w.r.t θ   ) to zero and solving for  θ . What you have is the MLE of  θ.

How do I compare a void pointer in C with another pointer?

In C, you can convert pointers to object types to void*. That in turn allows you to compare them for equality:void g(double *pd, void *pv) {
if (pv == (void*)pd) {
// ...
}
}
However, that doesn’t work for pointers to functions in C:void f(void (*pf)(), void *pv) {
if (pv == (void*)pf) { // Invalid C.
// ...
}
}
That said, many C compiler will accept that anyway.(Pointers-to-functions sometimes have interesting representations to deal with shared library schemes.)In standard C the pointer-to-function to void* conversion is definitely invalid though.In C++, on the other hand, that conversion is “conditionally supported”, which means that a compiler can reject the conversion, or it has to do the right thing with it. The C-style cast notation can be used, but it’s equivalent to a reinterpret_cast.