What happens if I assign to an int pointer a float variable address and vice-versa?

Question

I was wondering what would happen if I gave an int pointer a float variable address and vice-versa, so I tried it but couldn't exactly understand what's happening so if anyone can explain I would be grateful.

int n = 5;
float f = 1.21;
float *pf = &n;
int *pn = &f;

printf("%p  %p\n", pn, pf);
printf("%p  %p\n", &f, &n);
printf("%d   %f \n", *pn, *pf);
printf("%f   %d \n", n, f);
printf("%d  %f \n", n, f);

Output:

0xffffcba8  0xffffcbac
0xffffcba8  0xffffcbac
1067114824   0.000000
0.000000   0
5  1.210000

It's called **undefined behavior** - so no in general it can't be described. The only valid print statement is the last one. — Support Ukraine, Feb 11 '20 at 17:01
Also first and second print statements should be ok. After all they're just addresses. — Roberto Caboni, Feb 11 '20 at 17:16
@CmedCmbadi Please re-accept the [answer from chqrlie](https://stackoverflow.com/a/60174143/12139179), not mine. His one is way better. — RobertS supports Monica Cellio, Feb 11 '20 at 17:23

chqrlie · Answer 1 · 2020-02-11T17:27:05.387

3

There is a number of instances of undefined behavior in your code:

int n = 5; : OK
float f = 1.21; : OK
float *pf = &n; initializes a pointer with the value of a pointer to a different type. The C Standard makes no guarantees as to what this does nor as to what happens when you dereference this pointer. With -Wall -W, you would definitely get a warning for this, which you could silence with a cast: float *pf = (float *)&n;, yet the result is the same, dereferencing the pointer has undefined behavior.
int *pn = &f; same as above.
printf("%p %p\n", pn, pf); -> you should cast the pointers as (void *) because %p expects a void * as an argument. On some rare systems (old Cray systems IIRC), pointers to int, float and void may have different representations, causing this call to printf to behave in unexpected ways. Use this instead:
```
printf("%p  %p\n", (void *)pn, (void *)pf);
```
printf("%p %p\n", &f, &n); -> same problem. Use this instead:
```
printf("%p  %p\n", (void *)&f, (void *)&n);
```
The output is correct and the same for both expressions because your system is regular, but the C Standard does not guarantee it.
printf("%d %f \n", *pn, *pf); this one is correct but dereferencing the pointer has undefined behavior.
printf("%f %d \n", n, f); passing value of types different from the expected types has undefined behavior. The output you get is incorrect.
printf("%d %f \n", n, f); this one is correct and the output is 5 1.210000 as expected.

edited Feb 11 '20 at 17:27

answered Feb 11 '20 at 17:12

chqrlie

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@4386427 I´m a fair player. This answer is way better than mine. I´ll let the OP know. But this is no reason to downvote my answer. – RobertS supports Monica Cellio Feb 11 '20 at 17:21
1

@4386427 This is indeed the best answer. Nevertheless downvoting the accepted answer is a bit _harsh_. – Roberto Caboni Feb 11 '20 at 17:21
@RobertoCaboni Maybe "harsh" but not wrong. Consider that. – RobertS supports Monica Cellio Feb 11 '20 at 17:22
1

@RobertSsupportsMonicaCellio I am on your side. :) I was just trying to say that _downvoting your answer_ is harsh. ... because it is not wrong. – Roberto Caboni Feb 11 '20 at 17:23
@RobertoCaboni I´d misunderstood it, sorry. ;-) – RobertS supports Monica Cellio Feb 11 '20 at 17:26
1

Just in case anyone wonders, I agree and did not DV RobertS' answer :) – chqrlie Feb 11 '20 at 17:29
@chqrlie ;>) - =) – RobertS supports Monica Cellio Feb 11 '20 at 17:32

RobertS supports Monica Cellio · Accepted Answer · 2020-02-11T17:28:52.807

Chqrlie´s answer is way better than mine. Please re-accept his answer.

You should not abuse pointers and objects - Even not for curiosity reasons. The results of the 1., 2. and the 5. printf command are correct, but the results of the 3. and 4. printf command are products of Undefined Behavior, thus noone can explain them.

For more information about Undefined Behavior: Undefined, unspecified and implementation-defined behavior

What happens if I assign to an int pointer a float variable address and vice-versa?

2 Answers2