difference between setting up pointer with address-of operator

Question

I wondered what is the big difference between setting up C pointers in this way:

int int_var = 5;
int *int_ptr = &int_var;

And this way:

int int_var = 5;
int *int_ptr = int_var;

Since in both cases the result of *int_ptr will be 5, no?

Answer 1

No, only in the first case. The second case will cause undefined behavior when you'll try to deference the pointer. Use the first case.

Some explanation:

int int_var = 5;
int *int_ptr = &int_var; // here int_ptr will hold the address of var

Whereas

int int_var = 5;
int *int_ptr = int_var; // here int_ptr will hold the address 5.

Answer 2

No. In first case 5, in second case is undefined, is the conten of the memmory with adress 5. ??

Answer 3

No the int_ptr is a pointer so you have to assign an address to it when you define it

The first is the right one

int int_var = 5;
int *int_ptr = &int_var;

the other is wrong

the other one you are assigning an int value 5 to the pointer int_ptr (address) so it's like you assign 5 as address for the pointer int_ptr

Answer 4

In C, pointers point to the address itself. The address-of & operator is where the address of the variable is.

So in:

int int_var = 5;
int *int_ptr = &int_var;

*int_ptr would correctly be used since it is pointed to the ADDRESS of int_var which has the value of 5

However in:

int int_var = 5;
int *int_ptr = int_var;

*int_ptr points to an address that is 5, NOT the address where value 5 is located, which would be some random number.

In addition about arrays:

char *y;
char x[100];
y = x;

This can be done since an array is actually an address.

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