Exporting a default & declaring anon function all

Question

Here's an example of something I want to do and currently throws an error. I'm not sure I understand why, but it's not syntactically correct to export, assign default, and assign a variable on one line. The benefit of having it be an anon function is that i can use the fat arrow => and open up a return value with ( and ) instead of opening { and } to return jsx.

export default let Checkbox = (props) => (
  <div style={styles.checkboxContainer}>
    <input styleName={styles.checkbox} type="checkbox" />
    <span styleName={styles.checkboxStyled}></span>
  </div>
)

Is there a way to do this all in one line? Is there a good reason why I can't / why it's not in the spec?

Answer 1

You can do that for named exports if you want, but not for default ones.

Your alternatives are:

• Don't declare a variable. After all, you just want to export a single value:

  export default (props) => (
      …
  );
  

• Declare the variable and export it in two separate declarations:

  let Checkbox = (props) => (
      …
  );
  export default Checkbox;
  

  If you need to bind to the variable because you want to (re)assign to it (as let suggests), use

  export {Checkbox as default};
  

• Use a proper function declaration (which binds the local variable as well):

  export default function Checkbox(props) {
      return …
  }
  

  If you don't need that, you can also export an unnamed function declaration.