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Hi, does anyone know what is wrong with R. and R.call? While R.apply respects R., R.call does not. Here's an example:
const fn = R.curry((x, y) => x + y); const withApply = R.pipe( R.apply(R.__, ), R.apply(R.__, ), ); const withCall = R.pipe( R.call(R.__, 1), R.call(R.__, 2), ); withApply(fn); // returns 3, ok cool withCall(fn); // returns function (t,r) ...
Thing is, how can I make it work with R.call rather than R.apply?