AST 和 ASR 之间的区别¶
让我们看一个简单的 Fortran 代码：
integer function f(a, b) result(r)
integer, intent(in) :: a, b
integer :: c, d
c = a + b - d
r = c * a
end function
and look at what the AST and ASR look like.
AST¶
[1]:
%%showast
integer function f(a, b) result(r)
integer, intent(in) :: a, b
integer :: c, d
c = a + b - d
r = c * a
end function
(TranslationUnit
[(Function
f
[(a)
(b)]
[(AttrType
TypeInteger
[]
()
()
None
)]
r
()
()
[]
[]
[]
[(Declaration
(AttrType
TypeInteger
[]
()
()
None
)
[(AttrIntent
In
)]
[(a
[]
[]
()
()
None
())
(b
[]
[]
()
()
None
())]
()
)
(Declaration
(AttrType
TypeInteger
[]
()
()
None
)
[]
[(c
[]
[]
()
()
None
())
(d
[]
[]
()
()
None
())]
()
)]
[(Assignment
0
c
(- (+ a b) d)
()
)
(Assignment
0
r
(* c a)
()
)]
[]
[]
)]
)
AST 没有任何语义信息，但具有表示声明的节点，例如 integer, intent(in) :: a。诸如 a 之类的变量由 Name 节点表示，并且尚未连接到它们的声明。
The AST can also be exported in JSON, including source file name, line and column information: lfortran example.f90 --show-ast --json
ASR¶
[2]:
%%showasr
integer function f(a, b) result(r)
integer, intent(in) :: a, b
integer :: c, d
c = a + b - d
r = c * a
end function
(TranslationUnit
(SymbolTable
1
{
f:
(Function
(SymbolTable
2
{
a:
(Variable
2
a
[]
In
()
()
Default
(Integer 4)
()
Source
Public
Required
.false.
.false.
),
b:
(Variable
2
b
[]
In
()
()
Default
(Integer 4)
()
Source
Public
Required
.false.
.false.
),
c:
(Variable
2
c
[]
Local
()
()
Default
(Integer 4)
()
Source
Public
Required
.false.
.false.
),
d:
(Variable
2
d
[]
Local
()
()
Default
(Integer 4)
()
Source
Public
Required
.false.
.false.
),
r:
(Variable
2
r
[]
ReturnVar
()
()
Default
(Integer 4)
()
Source
Public
Required
.false.
.false.
)
})
f
(FunctionType
[(Integer 4)
(Integer 4)]
(Integer 4)
Source
Implementation
()
.false.
.false.
.false.
.false.
.false.
[]
.false.
)
[]
[(Var 2 a)
(Var 2 b)]
[(Assignment
(Var 2 c)
(IntegerBinOp
(IntegerBinOp
(Var 2 a)
Add
(Var 2 b)
(Integer 4)
()
)
Sub
(Var 2 d)
(Integer 4)
()
)
()
)
(Assignment
(Var 2 r)
(IntegerBinOp
(Var 2 c)
Mul
(Var 2 a)
(Integer 4)
()
)
()
)]
(Var 2 r)
Public
.false.
.false.
()
)
})
[]
)
ASR 有所有的语义信息（类型等），像 Function 这样的节点有一个符号表，没有任何声明节点。变量只是指向符号表的指针。
The ASR can also be exported in JSON, including source file name, line and column information: lfortran example.f90 --show-asr --json
讨论¶
上面是一个简单的例子。对于更复杂的示例，事情变得更加明显，例如：
integer function f2b(a) result(r)
use gfort_interop, only: c_desc1_int32
integer, intent(in) :: a(:)
interface
integer function f2b_c_wrapper(a) bind(c, name="__mod1_MOD_f2b")
use gfort_interop, only: c_desc1_t
type(c_desc1_t), intent(in) :: a
end function
end interface
r = f2b_c_wrapper(c_desc1_int32(a))
end function
AST 必须代表所有的 use 语句和 interface 块，并保持语义一致。
另一方面，ASR 跟踪符号表中的 c_desc1_int32、c_desc1_t 和 f2b_c_wrapper，并且知道它们是在 gfort_interop 模块中定义的，因此 ASR 没有任何这些声明节点。
从 ASR 转换为 AST 时，LFortran 将自动且正确地创建所有适当的 AST 声明节点。