Declare and initialize pointer concisely (i. e. pointer to int)

Question

Given pointers to char, one can do the following:

char *s = \"data\";

As far as I understand, a pointer variable is declared here, memory i

Answer 1

int i=42;
int *ptr = &i;

this is equivalent to writing

int i=42;
int *ptr;
ptr=&i;

Tough this is definitely confusing, but during function calls its quite useful as:

void function1()
{
int i=42;
function2(&i);
}

function2(int *ptr)
{
printf("%d",*ptr); //outputs 42
}

here, we can easily use this confusing notation to declare and initialize the pointer during function calls. We don't need to declare pointer globally, and the initialize it during function calls. We have a notation to do both at same time.

int *ptr; //declares the pointer but does not initialize it
//so, ptr points to some random memory location
*ptr=42; //you gave a value to this random memory location

Though this will compile, but it will invoke undefined behaviour as you actually never initialized the pointer.

Also,

char *ptr;
char str[6]="hello";
ptr=str;

EDIT: as pointed in the comments, these two cases are not equivalent. But pointer points to "hello" in both cases. This example is written just to show that we can initialize pointers in both these ways (to point to hello), but definitely both are different in many aspects.

char *ptr;
ptr="hello";

As, name of string, str is actually a pointer to the 0th element of string, i.e. 'h'. The same goes with any array arr[], where arr contains the address of 0th element.