组件管理 服务治理 … 权限 K8s Istio Envoy Tekton Argo KEDA ES InfluxDB Promethues Knative Ingress Rook Kube eventer … 策略 机制 Jaeger 实例 调度策略 链路 K8s及云原生生态给 开发者提供的是机制 开发者直接使用K8s的失败故事 • 认知成本高：K8

= array[Rows, Fig] Board = array[Cols, Col] var b: Board const a = 0 rook = 5 # whatever makes sense b[a][0] = rook echo b[a][0] # 5 # with user-defined templates we can simplify the index notation int): int8 = b[i][j] template `[]=`(b: var Board; i, j: int; v: int8) = b[i][j] = v b[a, 0] = rook echo b[a, 0] # 5 Now, let’s investigate the case where one or both dimensions of the matrix can

= array[Rows, Fig] Board = array[Cols, Col] var b: Board const a = 0 rook = 5 # whatever makes sense b[a][0] = rook echo b[a][0] # 5 # with user-defined templates we can simplify the index notation int): int8 = b[i][j] template `[]=`(b: var Board; i, j: int; v: int8) = b[i][j] = v b[a, 0] = rook echo b[a, 0] # 5 Now, let’s investigate the case where one or both dimensions of the matrix can

iteration julia> l == S.L && q == S.Q true LinearAlgebra.bunchkaufman – Func�on. bunchkaufman(A, rook::Bool=false; check = true) -> S::BunchKaufman Compute the Bunch-Kaufman 3 factoriza�on of a Symmetric the components S.D, S.U or S.L as appropriate given S.uplo, and S.p. If rook is true, rook pivo�ng is used. If rook is false, rook pivo�ng is not used. When check = true, an error is thrown if the decomposi�on julia> d == S.D && u == S.U && p == S.p true LinearAlgebra.bunchkaufman! – Func�on. bunchkaufman!(A, rook::Bool=false; check = true) -> BunchKaufman bunchkaufman! is the same as bunchkaufman, but saves space

iteration julia> l == S.L && q == S.Q true LinearAlgebra.bunchkaufman – Func�on. bunchkaufman(A, rook::Bool=false; check = true) -> S::BunchKaufman Compute the Bunch-Kaufman 3 factoriza�on of a Symmetric the components S.D, S.U or S.L as appropriate given S.uplo, and S.p. If rook is true, rook pivo�ng is used. If rook is false, rook pivo�ng is not used. When check = true, an error is thrown if the decomposi�on julia> d == S.D && u == S.U && p == S.p true LinearAlgebra.bunchkaufman! – Func�on. bunchkaufman!(A, rook::Bool=false; check = true) -> BunchKaufman bunchkaufman! is the same as bunchkaufman, but saves space

iteration julia> l == S.L && q == S.Q true LinearAlgebra.bunchkaufman – Func�on. bunchkaufman(A, rook::Bool=false; check = true) -> S::BunchKaufman Compute the Bunch-Kaufman 3 factoriza�on of a Symmetric the components S.D, S.U or S.L as appropriate given S.uplo, and S.p. If rook is true, rook pivo�ng is used. If rook is false, rook pivo�ng is not used. When check = true, an error is thrown if the decomposi�on julia> d == S.D && u == S.U && p == S.p true LinearAlgebra.bunchkaufman! – Func�on. bunchkaufman!(A, rook::Bool=false; check = true) -> BunchKaufman bunchkaufman! is the same as bunchkaufman, but saves space

iteration julia> l == S.L && q == S.Q true LinearAlgebra.bunchkaufman – Func�on. bunchkaufman(A, rook::Bool=false; check = true) -> S::BunchKaufman Compute the Bunch-Kaufman 3 factoriza�on of a Symmetric the components S.D, S.U or S.L as appropriate given S.uplo, and S.p. If rook is true, rook pivo�ng is used. If rook is false, rook pivo�ng is not used. When check = true, an error is thrown if the decomposi�on julia> d == S.D && u == S.U && p == S.p true LinearAlgebra.bunchkaufman! – Func�on. bunchkaufman!(A, rook::Bool=false; check = true) -> BunchKaufman bunchkaufman! is the same as bunchkaufman, but saves space

= array[Rows, Fig] Board = array[Cols, Col] var b: Board const a = 0 rook = 5 # whatever makes sense b[a][0] = rook echo b[a][0] # 5 # with user-defined templates we can simplify the index notation int): int8 = b[i][j] template `[]=`(b: var Board; i, j: int; v: int8) = b[i][j] = v b[a, 0] = rook echo b[a, 0] # 5 Now, let’s investigate the case where one or both dimensions of the matrix can grow

= array[Rows, Fig] Board = array[Cols, Col] var b: Board const a = 0 rook = 5 # whatever makes sense b[a][0] = rook echo b[a][0] # 5 # with user-defined templates we can simplify the index notation int): int8 = b[i][j] template `[]=`(b: var Board; i, j: int; v: int8) = b[i][j] = v b[a, 0] = rook echo b[a, 0] # 5 Now, let’s investigate the case where one or both dimensions of the matrix can grow

= array[Rows, Fig] Board = array[Cols, Col] var b: Board const a = 0 rook = 5 # whatever makes sense b[a][0] = rook echo b[a][0] # 5 # with user-defined templates we can simplify the index notation int): int8 = b[i][j] template `[]=`(b: var Board; i, j: int; v: int8) = b[i][j] = v b[a, 0] = rook echo b[a, 0] # 5 Now, let’s investigate the case where one or both dimensions of the matrix can grow