CODEWITHSUNDEEP
swiftbinaryintegeroverflowstandard-library
David Taylor
David Taylor

Reputation: 71

Why Int8.max &+ Int8.max equals to "-2"?

Following Swift Standard Library documentation, &+ discards any bits that overflow the fixed width of the integer type. I just did not get why adding two maximum values, 8-bit signed integer can hold results in -2:

/// Two max Int8 values (127 each, 8-bit group)
let x6 = Int8.max
let x7 = Int8.max

/// Prints `1 1 1 1 1 1 1`
String(Int8.max, radix: 2)

/// Here we get `-2` in decimal system
let x8 = x6 &+ x7

/// Prints `-1 0`
String(x8, radix: 2)

If we break down the binary calculation we will get this:

    1   1   1   1   1   1   1
+   1   1   1   1   1   1   1
-----------------------------
1   1   1   1   1   1   1   0

Which is -126, as the leftmost bit is a negative sign.

Why does Swift discards any bits except the rightmost two (1 and 0). Did I miss some overflow rules? I've read some pieces of knowledge in the web, but did not get closed to cracking this one.

Upvotes: 0

Views: 504

Answers (1)

Sweeper
Sweeper

Reputation: 272760

Swift (and every other programming language I know) uses 2's complement to represent signed integers, rather than sign-and-magnitude as you seem to assume.

In the 2's complement representation, the leftmost 1 does not represent "a negative sign". You can think of it as representing -128, so the Int8 value of -2 would be represented as 1111 1110 (-128 + 64 + 32 + 16 + 8 + 4 + 2).

OTOH, -126 would be represented as 1000 0010 (-128 + 2).

Upvotes: 0

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