Challenge 3: Why if (x = 5) Compiles, and Why Rust Can't Have This Bug — Possible Solution
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`if (x = 5)` compiles cleanly in C because of two facts combining: (1)
`=` is an assignment EXPRESSION in C, not just a statement -- an
assignment evaluates to the value that was assigned, so `x = 5` is
itself a valid expression whose value is 5; and (2) per the chapter,
`if` accepts ANY expression as its condition, evaluated as "is this
nonzero" -- there is no requirement that the condition be a genuine
comparison or a boolean at all.

What it actually does: `x = 5` assigns 5 to x (a real, permanent side
effect the programmer likely didn't intend), and the `if` then
evaluates the RESULT of that assignment (5) for truthiness. Since 5 is
nonzero, the condition is true -- always, regardless of whatever value
x held before. The programmer almost certainly meant `if (x == 5)` (a
comparison), but wrote `=` (assignment) by mistake, and C's rules allow
the mistake to compile as a different, silently valid program instead
of catching it as an error.

Why the equivalent mistake isn't possible in Rust: Rust's assignment
(`x = 5`) evaluates to the unit type `()`, not to the assigned value --
so even if someone wrote `if x = 5 { ... }`, the condition wouldn't be
usable as a boolean at all (`()` isn't a `bool`), and Rust's compiler
requires an `if` condition to be a genuine `bool` expression, full
stop. The combination that makes C's version dangerous -- assignment
producing a usable value, AND any nonzero value being an acceptable
condition -- simply doesn't exist in Rust; either piece alone would be
enough to block it, and Rust removes both.

WHY THIS WORKS AS AN ANSWER
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This identifies the TWO separate C language rules (assignment as a
value-producing expression, plus any-nonzero-condition) that combine to
allow the bug, rather than treating it as one vague quirk, and explains
precisely which of Rust's design choices removes each half of that
combination.