Why does derefrencing a raw pointer gives a segmentation fault if there is no shared reference to the value pointed at?

user_13751942 0
Viewed 67

Running the following code gives a segmentation fault:

fn main() {
    let val = 1;
    let ptr = val as *const i32;
    unsafe { println!("{:?}", *ptr) };
}

Output:

[1]    69212 segmentation fault (core dumped)  cargo r

However, when val is put in as a reference & while declaring the raw pointer, the code runs as intended and as val is printed out.

fn main() {
    let val = 1;
    let ptr = &val as *const i32;
    unsafe { println!("{:?}", *ptr) };
}

Output:

1

So what is the shared reference doing here and why does the program fail without it? Isn't a reference in rust also a pointer with extra schematics? Why to we need to create a pointer to a reference and not directly to the val itself?

raw-pointer rust pointers memory
1 Answers

This issue can be answered by looking at the different semantics of the both code lines you provided.

fn main() {
    let val = 1;
    println!("{:?}", val as *const i32); // Output: 0x1
    println!("{:?}", &val as *const i32); // Output: 0x7ff7b36a4eec (probably little different)
}

Without the reference the value of the variable is take as it is to be used to dereference the memory. This leads of course to a segmentation fault, since it will be not in the allowed address range of the program.

Only when the reference operator is used, the address of the variable is casted to a raw pointer, which then later can be dereferenced without any segmentation fault.

user_11665930
Related