Why it gives segfault?

Question

#include <iostream>
int main(){
int a{45};
double* pd=reinterpret_cast<double*>(a);
std::cout<<*pd<<std::endl;
return 0;
}

Answer 1

I guess you are not allowed to access memory address 45.

Perhaps you meant to reinterpret the value 45 as a double. To fix it, replace (a) with (&a).

Answer 2

Also, note that you have a buffer overflow error and you should use `long` for `a` to be the same size as `double`. Otherwise it uses extra random data nearby (in stack) and gives different results for each run.

Answer 3

reinterpret_cast is undefined territory and should be avoided.
Good to know.
But even better to understand the reason to not use it unless absolutely necessary.

https://stackoverflow.com/questions/573294/when-to-use-reinterpret-cast

nice discussion with lots of examples.

Answer 4

Ujjawal Gupta
Ok got it
Keep that curiosity up on fire : )

Answer 5

Why go through all these to have a `double` value from `int` value? what is your plan?

I'm sure you know these types has incompatibilities, and that inter-type casting *may* introduce a flaw (can't expect reliability on precision)

Answer 6

Ipang ikr, casting is an ugly and dangerous operation, but actually I wants to learn how reinterpret_cast works

Answer 7

Ipang no, there's no such type. Maybe they meant "a float with no sign", i. e. non-negative?

Answer 8

Евгений
Yes also IIRC signedness modifier `unsigned` is only for integral types. I find it hard to believe nobody noticed that, in a strict place like stackoverflow ...

Answer 9

I edited the answer at stackoverflow, now there's no "unsigned float" nonsense.

Answer 10

Brian I think it has to do something with bit level representation of int and double 🤔

Answer 11

Ipang sure :)

Answer 12

Bob_Li gotcha

Answer 13

Bob_Li
I can't help to wonder if it is true there is `unsigned float` type as someone mentioned in one of the answers in the thread you attached ...

https://stackoverflow.com/a/43273907

"When you convert for example int(12) to unsigned float (12.0f) your processor needs to invoke some calculations as both numbers has different bit representation ..."

Is there really `unsigned float` type?

Answer 14

Ipang
unsigned float... I didn't notice that, too.😁 Good catch. I never wondered about why there is no unsigned float or double. Now I know.
https://stackoverflow.com/questions/2523025/unsigned-double-in-c

Answer 15

Thanks for the edit Евгений

From now on people who follows up on that thread in stackoverflow (and this one here) are saved from a misunderstanding possibly arised from the `unsigned float` typo : )

Answer 16

Евгений Roger that