# Traps for Young Players

This section describes the two most common sources of confusion encountered when using Magma’s evaluation strategy.

## Trap 1

We boot Magma. It begins with an initial context something like `[ ..., (’+’,A), (’-’,S), ... ]` where $A$ is the (function) value that is the addition function, and $S$ is the (function) value that is the subtraction function. Now say we type,

```
> '+' := '-';
> 1 + 2;

```

Magma will respond with the answer `-1`. To see why this is so consider the effect of each line on the current context. After the first line the current context will be `[ ..., (’+’,S), (’-’,S), ... ]`, where S is as before. The identifier `+` has been re-assigned. Its new value is the value of the identifier ‘-’ in the current context, and the value of ‘-’ is the (function) value that is the subtraction function. Hence in the second line when Magma replaces the identifier `+` with its value in the current context, the value that is substituted is therefore S, the subtraction function!

## Trap 2

Say we type,

```
> f := func< n | n + 1 >;
> g := func< m | m + f(m) >;

```

After the first line the current context is `[ (f,FUNC( n : n+1)) ]`. After the second line the current context is `[ (f,FUNC( n : n+1)), (g,FUNC(m : m + FUNC(n : n+1)(m))) ]`. If we now type,

```
> g(6);

```

Magma will respond with the answer 13. Now say we decide that our definition of $f$ is wrong. So we now type in a new definition for $f$ as follows,

```
> f := func< n | n + 2 >;

```

If we again type,

```
> g(6);

```

Magma will again reply with the answer 13! To see why this is so consider how the current context changes. After typing in the initial definitions of $f$ and $g$ the current context is `[ (f, FUNC(n : n+1)), (g, FUNC(m : m + FUNC(n : n+1)(m))) ]`. After typing in the second definition of $f$ the current context is `[ (f, FUNC(n : n+2)), (g, FUNC(m : m + FUNC(n : n+1)(m)))]`. Remember that changing the *value* of one identifier, in this case $f$, does *not* change the value of any other identifiers, in this case $g$! In order to change the value of $g$ to reflect the new value of $f$, $g$ would have to be re-assigned.