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math/big: use built-in clear to simplify code
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parent
f1d60500bc
commit
2ba8c4c705
3 changed files with 16 additions and 24 deletions
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@ -533,10 +533,8 @@ func (x *Int) Bytes() []byte {
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//
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// If the absolute value of x doesn't fit in buf, FillBytes will panic.
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func (x *Int) FillBytes(buf []byte) []byte {
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// Clear whole buffer. (This gets optimized into a memclr.)
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for i := range buf {
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buf[i] = 0
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}
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// Clear whole buffer.
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clear(buf)
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x.abs.bytes(buf)
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return buf
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}
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@ -44,12 +44,6 @@ func (z nat) String() string {
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return "0x" + string(z.itoa(false, 16))
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}
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func (z nat) clear() {
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for i := range z {
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z[i] = 0
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}
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}
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func (z nat) norm() nat {
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i := len(z)
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for i > 0 && z[i-1] == 0 {
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@ -196,7 +190,7 @@ func (z nat) mulAddWW(x nat, y, r Word) nat {
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// basicMul multiplies x and y and leaves the result in z.
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// The (non-normalized) result is placed in z[0 : len(x) + len(y)].
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func basicMul(z, x, y nat) {
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z[0 : len(x)+len(y)].clear() // initialize z
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clear(z[0 : len(x)+len(y)]) // initialize z
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for i, d := range y {
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if d != 0 {
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z[len(x)+i] = addMulVVW(z[i:i+len(x)], x, d)
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@ -222,7 +216,7 @@ func (z nat) montgomery(x, y, m nat, k Word, n int) nat {
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panic("math/big: mismatched montgomery number lengths")
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}
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z = z.make(n * 2)
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z.clear()
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clear(z)
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var c Word
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for i := 0; i < n; i++ {
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d := y[i]
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@ -443,8 +437,8 @@ func (z nat) mul(x, y nat) nat {
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y0 := y[0:k] // y0 is not normalized
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z = z.make(max(6*k, m+n)) // enough space for karatsuba of x0*y0 and full result of x*y
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karatsuba(z, x0, y0)
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z = z[0 : m+n] // z has final length but may be incomplete
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z[2*k:].clear() // upper portion of z is garbage (and 2*k <= m+n since k <= n <= m)
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z = z[0 : m+n] // z has final length but may be incomplete
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clear(z[2*k:]) // upper portion of z is garbage (and 2*k <= m+n since k <= n <= m)
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// If xh != 0 or yh != 0, add the missing terms to z. For
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//
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@ -497,7 +491,7 @@ func basicSqr(z, x nat) {
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n := len(x)
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tp := getNat(2 * n)
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t := *tp // temporary variable to hold the products
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t.clear()
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clear(t)
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z[1], z[0] = mulWW(x[0], x[0]) // the initial square
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for i := 1; i < n; i++ {
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d := x[i]
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@ -592,7 +586,7 @@ func (z nat) sqr(x nat) nat {
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z = z.make(max(6*k, 2*n))
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karatsubaSqr(z, x0) // z = x0^2
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z = z[0 : 2*n]
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z[2*k:].clear()
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clear(z[2*k:])
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if k < n {
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tp := getNat(2 * k)
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@ -723,7 +717,7 @@ func (z nat) shl(x nat, s uint) nat {
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n := m + int(s/_W)
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z = z.make(n + 1)
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z[n] = shlVU(z[n-m:n], x, s%_W)
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z[0 : n-m].clear()
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clear(z[0 : n-m])
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return z.norm()
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}
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@ -769,7 +763,7 @@ func (z nat) setBit(x nat, i uint, b uint) nat {
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case 1:
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if j >= n {
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z = z.make(j + 1)
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z[n:].clear()
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clear(z[n:])
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} else {
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z = z.make(n)
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}
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@ -734,7 +734,7 @@ func (z nat) divRecursive(u, v nat) {
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tmp := getNat(3 * len(v))
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temps := make([]*nat, recDepth)
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z.clear()
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clear(z)
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z.divRecursiveStep(u, v, 0, tmp, temps)
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// Free temporaries.
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@ -758,7 +758,7 @@ func (z nat) divRecursiveStep(u, v nat, depth int, tmp *nat, temps []*nat) {
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u = u.norm()
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v = v.norm()
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if len(u) == 0 {
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z.clear()
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clear(z)
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return
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}
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@ -816,7 +816,7 @@ func (z nat) divRecursiveStep(u, v nat, depth int, tmp *nat, temps []*nat) {
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// Compute the 2-by-1 guess q̂, leaving r̂ in uu[s:B+n].
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qhat := *temps[depth]
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qhat.clear()
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clear(qhat)
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qhat.divRecursiveStep(uu[s:B+n], v[s:], depth+1, tmp, temps)
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qhat = qhat.norm()
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@ -833,7 +833,7 @@ func (z nat) divRecursiveStep(u, v nat, depth int, tmp *nat, temps []*nat) {
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// But we can do the subtraction directly, as in the comment above
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// and in long division, because we know that q̂ is wrong by at most one.
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qhatv := tmp.make(3 * n)
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qhatv.clear()
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clear(qhatv)
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qhatv = qhatv.mul(qhat, v[:s])
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for i := 0; i < 2; i++ {
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e := qhatv.cmp(uu.norm())
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@ -864,11 +864,11 @@ func (z nat) divRecursiveStep(u, v nat, depth int, tmp *nat, temps []*nat) {
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// Choose shift = B-1 again.
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s := B - 1
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qhat := *temps[depth]
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qhat.clear()
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clear(qhat)
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qhat.divRecursiveStep(u[s:].norm(), v[s:], depth+1, tmp, temps)
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qhat = qhat.norm()
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qhatv := tmp.make(3 * n)
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qhatv.clear()
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clear(qhatv)
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qhatv = qhatv.mul(qhat, v[:s])
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// Set the correct remainder as before.
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for i := 0; i < 2; i++ {
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