What is the difference between the ways of declaring a pointer? [duplicate]

What changes between these two ways?

int *ptr and int* ptr

Both work the same way, but for my code or performance my program has difference?

#include <iostream>

using namespace std;

int main() {

    int a(5);
    int *ptr1 = &a;
    int* ptr2 = &a;

    cout << "a:    (0x" << hex << uppercase << (uintptr_t)&a << ") = " << dec << a << endl;
    cout << "ptr1: (0x" << hex << uppercase << (uintptr_t)ptr1 << ") = " << dec << *ptr1 << endl;
    cout << "ptr2: (0x" << hex << uppercase << (uintptr_t)ptr2 << ") = " << dec << *ptr2 << endl;
    
    return 0;
}

Output:

a:    (0x61FE0C) = 5
ptr1: (0x61FE0C) = 5
ptr2: (0x61FE0C) = 5

Absolutely nothing. Whitespace is ignored.

Where to align is a decision of style, based on team conventions and rules (if any) or personal preference, and is mostly to improve the readability of the code. It doesn't change how the code is compiled.

They're the same:

int a {5};
int *ptr1 = &a; // same as below
int* ptr2 = &a;

Not just for pointers but also for references:

int a {5};
int& b = a;
int &c = b; // same as above
int & d = a; // same

The both declarations have the same meaning. You could consider even two more declarations

int *ptr1 = &a;
int* ptr2 = &a;
int * ptr3 = &a;
int*ptr4 = &a;

or

int*ptr4=&a;

The compiler can parse these records correctly without an ambiguity. The symbol * may be a part neither of an identifier nor of a keyword.

The difference between the records of these two declarations

int *ptr1 = &a;
int* ptr2 = &a;

lies in the semantic between declarators and type specifiers.

For example you may write

int ( *ptr ) = &a;

but you may not write

( int * ) ptr = &a;

Or you may write

int typedef *ptr;

but you may not write

int * typedef ptr;

So it is preferable to write

int *ptr1 = &a;

instead of

int* ptr2 = &a;

For example this declaration

int* ptrq, ptr2;

does not declare two pointers. Instead you need to write

int *ptr1, *ptr2;

Or if you have a pointer to an array like

int a[10];

then you even can not use the combination int*. You have to write

int ( *ptr )[10] = &a;