# Vocabulary/ampco

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`[x] u&:v y`Appose Conjunction

Rank Infinity *-- operates on [x and] y as a whole --*
WHY IS THIS IMPORTANT?

Applies verb `v` to each argument **in its entirety**, and then applies verb `u` to the result(s) of `v`

u =: < NB. x is smaller than y v =: | NB. magnitude csm =: +/ @: u &: v NB. Count how many x's are smaller than y in magnitude _1 2 0 csm 2 _3 0 2

- Verb
`v`is executed monadically. - Verb
`u`is executed either monadically or dyadically depending whether`u&:v`has been called monadically or dyadically .

See: **More Information** for a visual comparison of At (`@:`), Atop (`@`), Compose (`&`) and Appose (`&:`).

### Common Uses

**The monadic use of**Some compounds of the form`&:`is deprecated. Use`@:`instead.`f&:g`are not recognized for special code in places where`f@:g`is recognized, if only the monadic form of the compounds is eligible for special treatment.`u&:v`is called for when the rank of`v`is less than the ranks of an argument, but you want to apply`u`to the**entire result**of`v`.

In the `csm` example above, we needed (`&:`) not (`&`)

_1 2 0 +/@:<&:| 2 _3 0 NB. "csm" example as an anonymous verb in a 1-liner 2 _1 2 0 +/@:<&| 2 _3 0 NB. different result using (&) in place of (&:) 1 1 0

Because ` | y ` has rank `0`, the entire compound ` +/@:<&| ` is applied to each *atom* of `x` and `y`, making the `+/` useless since it now operates on each atom individually.

### Related Primitives

Atop (`@`),
At (`@:`),
Compose (`&`),
Hook (`(u v)`),
Fork (`(f g h)`)

### More Information

1. Contrast **Appose** (`u&:v`) with Compose (`u&v`) which applies `v` to **individual cell(s)** and then applies `u` on the **individual result(s)** of `v`

2. The difference between **Appose** (`u&:v`) and Compose (`u&v`) is shown in the **last two** columns of the diagram below:

The above diagram also shows At (`@:`) and Atop (`@`) for overall comparison.

3. So what's the difference between Atop (`@`) and Compose (`&`) ?

None at all, for the *monads* (`u@v`) and (`u&v`)

u&v y ↔ u v y u@v y ↔ u v y

But thedyadsare different

x u&v y ↔ (v x) u (v y) x u@v y ↔ u x v y

According to the J Dictionary -- `&:` *is equivalent to* `&` *except that the ranks of the resulting function are infinite; the relation is similar to that between* `@:` *and* `@`