= Solution
Accumulate the prefix by <Horner's method> in base two. `digit-'0'` converts either permitted digit to its numerical value:
``
var i = 0
var integerPart = 0.0
while (i < bin.length && bin(i) != '.') {
integerPart = 2.0 * integerPart + (bin(i) - '0')
i += 1
}
``
The <loop invariant>, in exact arithmetic, is that `i` is the number of processed digits before the point and
$$
integerPart=\sum_{j=0}^{i-1}(bin(j)-\mathtt{'0'})2^{i-1-j}.
$$
Appending a digit shifts the old positional value by one binary place and adds that digit, which proves preservation. The <loop variant> is $bin.length-i$. At exit, `i` is the binary-point index or the length if no point occurs, and \b[`integerPart` is the value of the <integer> prefix]. An empty prefix contributes zero. There is one multiplication per processed digit, with no separate powers or repeated divisions. <Short-circuit evaluation> prevents an out-of-range access when the point is absent.
For `Double`, this exact-arithmetic <invariant> describes the intended value; the operational value is the rounded Horner recurrence. Very long prefixes can lose precision or overflow, as discussed in part (g). A fixed-width return type cannot represent every finite binary digit <string> exactly.
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