babeljs - 我可以防止Babel遍历插件插入的代码吗？

Question

我正在构建一个插件，该插件通过调用enterFunction()在每个现有函数调用的前面插入path.insertBefore。所以我的代码是从:

myFunction();

至:

enterFunction();
myFunction();

问题是，当我插入节点时，Babel再次遍历插入的节点。这是日志记录输出:

'CallExpression', 'myFunction'
'CallExpression', 'enterFunction'

如何防止Babel输入 enterFunction调用表达式及其子级？

这是我当前用于Babel插件的代码:

function(babel) {
    return {
        visitor: {
            CallExpression: function(path) {
                console.log("CallExpression", path.node.callee.name)
                if (path.node.ignore) {
                    return;
                }
                path.node.ignore = true

                var enterCall = babel.types.callExpression(
                    babel.types.identifier("enterFunction"), []
                )
                enterCall.ignore = true;
                path.insertBefore(enterCall)
            }
        }
    }
}

Answer 1

Babel Handbook提到以下部分:

If your plugin needs to not run in a certain situation, the simpliest thing to do is to write an early return.

BinaryExpression(path) {
  if (path.node.operator !== '**') return;
}

If you are doing a sub-traversal in a top level path, you can use 2 provided API methods:

path.skip() skips traversing the children of the current path. path.stop() stops traversal entirely.

outerPath.traverse({
  Function(innerPath) {
    innerPath.skip(); // if checking the children is irrelevant
  },
  ReferencedIdentifier(innerPath, state) {
    state.iife = true;
    innerPath.stop(); // if you want to save some state and then stop traversal, or deopt
  }
});

简而言之，请使用 path.skip()跳过遍历当前路径的子级。
this snippet using Visitors, CallExpression and skip()说明了此方法的一种应用:

export default function (babel) {
  const { types: t } = babel;

  return {
    name: "ast-transform", // not required
    visitor: {
      CallExpression(path) {
        path.replaceWith(t.blockStatement([
          t.expressionStatement(t.yieldExpression(path.node))
        ]));
        path.skip();
      }
    }
  };
}