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cesarb
on Nov 23, 2017
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Announcing Rust 1.22
In C, the value of x in that fragment is still the same: a pointer (ok, not exactly, but something which behaves like a pointer) to a memory location where an int can be stored. The f(x) call can't change that pointer.
int_19h
on Nov 23, 2017
[–]
Try doing something like &x, and you'll see that it doesn't actually reliably behave like a pointer.
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