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?

In the second case, `*int_ptr` will be whatever is stored at memory address 5. In other words, garbage. — jpm, Apr 09 '13 at 19:13
Hint: turn up the warning level on your compiler and/or pay attention to the warning you're already getting ;). — FatalError, Apr 09 '13 at 19:14
Do you use `gcc`? Then activate all warnings: `-Wall -Wextra -pedantic`. This should already tell you that there is something smelly going on. — Zeta, Apr 09 '13 at 19:14
@jpm It's more likely the OS will catch it as an invalid access since such low addresses are rarely mapped. — cnicutar, Apr 09 '13 at 19:14
@cnicutar In a sane run time, probably, but it's not guaranteed, so a crash (probably segfault) will be the best case scenario upon a dereference. — jpm, Apr 09 '13 at 19:17

score 2 · Accepted Answer · answered Apr 09 '13 at 19:13

2

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.

answered Apr 09 '13 at 19:13

Avidan Borisov

Ok, does this apply to arrays as well? – rel-s Apr 09 '13 at 19:13
@arielschon12 What do you mean? – Avidan Borisov Apr 09 '13 at 19:14
Should i write code like this: `int array[20]; int *ptr=array;` or with the address of operator? – rel-s Apr 09 '13 at 19:15
1

@arielschon12: You _can_ write code like this, because `int[20]` can decay to `int*`. You _shouldn't_, as long as you don't know what you're actually doing. – Zeta Apr 09 '13 at 19:16
@arielschon12 Depends on what you're trying to achieve. It won't copy anything (if that's what you expect to happen). – Avidan Borisov Apr 09 '13 at 19:16
@Zeta It is perfectly fine to write code like this for purposes such as iterating over an array. – Avidan Borisov Apr 09 '13 at 19:17

score 0 · Answer 2 · answered Apr 09 '13 at 19:14

0

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

answered Apr 09 '13 at 19:14

qPCR4vir

score 0 · Answer 3 · answered Apr 09 '13 at 19:15

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

score 0 · Answer 4 · edited Feb 09 '14 at 05:52

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.

4 Answers4