Spread operator in method shorthand in object literal is not emitted with "__assign"

[kobiburnley]

TypeScript Version: 2.4.0 / nightly (2.5.0-dev.201xxxxx)

Link to playground

Code

const y = {
  prop: "ms"
}

const x = {
  ...y,
  fn(options) {
    return {
      timestamps: false,
      ...options
    }
  }
}

Emits to:

var __assign = (this && this.__assign) || Object.assign || function(t) {
    for (var s, i = 1, n = arguments.length; i < n; i++) {
        s = arguments[i];
        for (var p in s) if (Object.prototype.hasOwnProperty.call(s, p))
            t[p] = s[p];
    }
    return t;
};
var y = {
    prop: "ms"
};
var x = __assign({}, y, { fn: function (options) {
        return {
            timestamps: false,
            ...options
        };
    } });

Expected behavior:
"...options" will be emitted with "__assign"
Actual behavior:
"...options" is not emitted with "__assign"

[aminpaks]

I think it's supposed to be:

const y = {
  prop: "ms"
}

const x = {
  ...y,
  ...((options) => {
    return {
      timestamps: false,
      ...options
    }
  })({new: 'old',})
}

or

const y = {
  prop: "ms"
}

const x = {
  ...y,
  ...fn({}),
}

function fn(options) {
  return {
    timestamps: false,
    ...options
  }
}

[kobiburnley]

The issue happens when using method shorthand in object literal

[aluanhaddad]

This seems to happen when their is a spread in the directly enclosing literal only.

This works

const x = {
  fn() {
    return {
      ...{}
    };
  }
}

and this fails

const x = {
  ...{},
  fn() {
    return {
      ...{}
    };
  }
}

but this works

const x = {
  ...{},
  y: {
    fn() {
      return {
        ...{}
      };
    }
  }
}

[aluanhaddad]

I just realized this is a dupe of #16765

[mhegazy]

Automatically closing this issue for housekeeping purposes. The issue labels indicate that it is unactionable at the moment or has already been addressed.