あっても(なくても)いいtypename

C++ Grammar
  • The keyword typename shall only be applied to qualified names, but those names need not be dependent. (14.6/5)
  • 上の文は、最新のDraftでは(typenameのルールを緩くするため？)削除されているようだ

[]struct[] []X[]

[]{[]

    []typedef[] []int[] []type[][];[]

[]};[]


[]template[][]< [][]class[] []T[][] >[]

[]typename[] []X[][]::[][]type[]

[]foo[][]([][]typename[] []X[][]::[][]type[] []a[][])[]

[]{[]

    []typename[] []X[][]::[][]type[] []b[][] = [][]0[][];[]

    []return[] []b[][];[]

[]}[]
[]template[][]< [][]class[] []T[][] >[]

[]X[][]::[][]type[]

[]bar[][]([][]X[][]::[][]type[] []a[][])[]

[]{[]

    []X[][]::[][]type[] []b[][] = [][]0[][];[]

    []return[] []b[][];[]

[]}[]
[]int[] []main[][]()[]

[]{[]

[]    ::[][]foo[][]<[][]double[][]>([][]1[][]);[]

[]    ::[][]bar[][]<[][]double[][]>([][]1[][]);[]

[]}[]