Update dart.md Fix bug in asynchronous login example #814
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当前行为
当前的异步登录示例代码存在以下问题:
main
函数中尝试使用await userName()
,但userName
是一个字符串变量,而不是一个函数。login
函数并使用await
来等待其结果。预期行为
main
函数应该正确地调用login
函数,并使用await
来等待其结果,然后打印出用户名。示例代码
以下是修正后的代码: