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std.mulWide() whose return is twice as wide

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This commit is contained in:
Shawn Landden 2019-03-27 15:45:18 -05:00 committed by Andrew Kelley
parent 64b2cf776c
commit 85d1885375
2 changed files with 12 additions and 8 deletions
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11
std/math.zig
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@ -806,3 +806,14 @@ test "max value type" {
const x: u32 = maxInt(i32);
testing.expect(x == 2147483647);
}
pub fn mulWide(comptime T: type, a: T, b: T) @IntType(T.is_signed, T.bit_count * 2) {
const ResultInt = @IntType(T.is_signed, T.bit_count * 2);
return ResultInt(a) * ResultInt(b);
}
test "math.wideMul" {
testing.expect(wideMul(u8, 5, 5) == 25);
testing.expect(wideMul(i8, 5, -5) == -25);
testing.expect(wideMul(u8, 100, 100) == 10000);
}

9
std/math/big/int.zig
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@ -705,14 +705,7 @@ pub const Int = struct {
const c1: Limb = @boolToInt(@addWithOverflow(Limb, a, carry.*, &r1));
// r2 = b * c
//
// We still use a DoubleLimb here since the @mulWithOverflow builtin does not
// return the carry and lower bits separately so we would need to perform this
// anyway to get the carry bits. The branch on the overflow case costs more than
// just computing them unconditionally and splitting.
//
// This could be a single x86 mul instruction, which stores the carry/lower in rdx:rax.
const bc = DoubleLimb(b) * DoubleLimb(c);
const bc = DoubleLimb(math.mulWide(Limb, b, c));
const r2 = @truncate(Limb, bc);
const c2 = @truncate(Limb, bc >> Limb.bit_count);