What does it mean when we equals content of a pointer variable to address of 1st element of array?

Question

int value[20];
int * ptr;
*ptr==&value[0];
what does the above code mean?
don't worry about the function.
it's quite different from
ptr=&value[0];

Answer 1

The statement *ptr==&value[0]; is an invalid statement, so it really means nothing. You can't compare an integer value (*ptr) to an address/int* in this case (&value[0]).
I created some code for you to look at in case it may help you understand.
https://code.sololearn.com/cAn4ufOuAkKX

if your intention is to point the pointer to the first element of the array, you first need to make sure you're using the assignment operator(=). The following two statements would do this:
ptr = value; //an array name is itself a pointer which is why this works
ptr = &value[0];