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Python symbol.simple_stmt方法代码示例

本文整理汇总了Python中symbol.simple_stmt方法的典型用法代码示例。如果您正苦于以下问题:Python symbol.simple_stmt方法的具体用法?Python symbol.simple_stmt怎么用?Python symbol.simple_stmt使用的例子?那么恭喜您, 这里精选的方法代码示例或许可以为您提供帮助。您也可以进一步了解该方法所在symbol的用法示例。


在下文中一共展示了symbol.simple_stmt方法的5个代码示例,这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞,您的评价将有助于系统推荐出更棒的Python代码示例。

示例1: single_input

# 需要导入模块: import symbol [as 别名]
# 或者: from symbol import simple_stmt [as 别名]
def single_input(self, node):
        ### do we want to do anything about being "interactive" ?

        # NEWLINE | simple_stmt | compound_stmt NEWLINE
        n = node[0][0]
        if n != token.NEWLINE:
            return self.com_stmt(node[0])

        return Pass() 
开发者ID:IronLanguages,项目名称:ironpython2,代码行数:11,代码来源:transformer.py

示例2: simple_stmt

# 需要导入模块: import symbol [as 别名]
# 或者: from symbol import simple_stmt [as 别名]
def simple_stmt(self, nodelist):
        # small_stmt (';' small_stmt)* [';'] NEWLINE
        stmts = []
        for i in range(0, len(nodelist), 2):
            self.com_append_stmt(stmts, nodelist[i])
        return Stmt(stmts) 
开发者ID:IronLanguages,项目名称:ironpython2,代码行数:8,代码来源:transformer.py

示例3: suite

# 需要导入模块: import symbol [as 别名]
# 或者: from symbol import simple_stmt [as 别名]
def suite(self, nodelist):
        # simple_stmt | NEWLINE INDENT NEWLINE* (stmt NEWLINE*)+ DEDENT
        if len(nodelist) == 1:
            return self.com_stmt(nodelist[0])

        stmts = []
        for node in nodelist:
            if node[0] == symbol.stmt:
                self.com_append_stmt(stmts, node)
        return Stmt(stmts)

    # --------------------------------------------------------------
    #
    # EXPRESSION NODES  (invoked by com_node())
    # 
开发者ID:IronLanguages,项目名称:ironpython2,代码行数:17,代码来源:transformer.py

示例4: com_node

# 需要导入模块: import symbol [as 别名]
# 或者: from symbol import simple_stmt [as 别名]
def com_node(self, node):
        # Note: compile.c has handling in com_node for del_stmt, pass_stmt,
        #       break_stmt, stmt, small_stmt, flow_stmt, simple_stmt,
        #       and compound_stmt.
        #       We'll just dispatch them.
        return self._dispatch[node[0]](node[1:]) 
开发者ID:IronLanguages,项目名称:ironpython2,代码行数:8,代码来源:transformer.py

示例5: get_docstring

# 需要导入模块: import symbol [as 别名]
# 或者: from symbol import simple_stmt [as 别名]
def get_docstring(self, node, n=None):
        if n is None:
            n = node[0]
            node = node[1:]
        if n == symbol.suite:
            if len(node) == 1:
                return self.get_docstring(node[0])
            for sub in node:
                if sub[0] == symbol.stmt:
                    return self.get_docstring(sub)
            return None
        if n == symbol.file_input:
            for sub in node:
                if sub[0] == symbol.stmt:
                    return self.get_docstring(sub)
            return None
        if n == symbol.atom:
            if node[0][0] == token.STRING:
                s = ''
                for t in node:
                    s = s + eval(t[1])
                return s
            return None
        if n == symbol.stmt or n == symbol.simple_stmt \
           or n == symbol.small_stmt:
            return self.get_docstring(node[0])
        if n in _doc_nodes and len(node) == 1:
            return self.get_docstring(node[0])
        return None 
开发者ID:IronLanguages,项目名称:ironpython2,代码行数:31,代码来源:transformer.py


注:本文中的symbol.simple_stmt方法示例由纯净天空整理自Github/MSDocs等开源代码及文档管理平台,相关代码片段筛选自各路编程大神贡献的开源项目,源码版权归原作者所有,传播和使用请参考对应项目的License;未经允许,请勿转载。