5

竜の歯を依り代とした人型はコルキス王の魔術と聞く。 その娘、王女メディアは稀代の魔女と謳われたそうだが?」

Does this mean:

The human form who made the dragon tooth as a vessel

or

A human form which was a dragon tooth made vessel.

How does にする・とする work before a name?

  • 1
    Related: japanese.stackexchange.com/q/18593/5010 – naruto Apr 22 '16 at 8:08
  • I found that, but my question here is if :XをYとするN means: "N makes X into Y" or if it means "N is X made into Y". – Splikie Apr 22 '16 at 8:14
  • 1
    As you can see in the linked question, XをYとする/したN is simply "N where X is Y" or "N with X as Y". BTW what's 人型? Is that a type of magic in this novel? – naruto Apr 22 '16 at 8:19
  • Yes, I read it a long time ago so I don't remember what it was specifically. It was not that important since it was never mentioned again. – Splikie Apr 22 '16 at 8:26
  • @naruto I have another question. If I wanted to say N makes X into Y I should say (XをYにさせるN) – Splikie Apr 22 '16 at 9:22
3

A を B に/とする means either make A into B or have A as B. Whether the change of state is involved or not (i.e. A is "made" into B or originally so) is unspecified and always left ambiguous, so you can only choose the right side from context.

人型 that made a dragon tooth into the (spirit) vessel
or
人型 using a dragon tooth as the (spirit) vessel

(I don't have any idea what 人型 stands for, though...)

This type of ambiguity is parallel to that of English passive construction.

The project is finally completed. → changes from "WIP" to "done".
The project is already completed. → keeps the "done" state.

If you dig a little deeper, using the particle に or と makes some difference in nuance. According to this page, と suggests the result is authentic, finalized, or established, whereas に comparatively ad-hoc, questionable, or figurative.

There are some idioms exclusively allow either of them.

身を粉にする
× 身を粉とする

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.