Anonymous function arguments and parameters extrction

If I have the anonymous function:

pu = @(x, y) [x,y];

is it npossible to extract the value of the argument x after the function has acted on it?

(Apologies if this does not make sense)

Answer 1

[Steven Lord]

As written, as long as you know how long either one of the inputs was, yes.

P = @(x, y) [x, y];
x1 = 1:5;
y1 = 6:10;
z = P(x1, y1)

z = 1×10

1 2 3 4 5 6 7 8 9 10

x2 = z(1:5); % I know how long x was, extract that many elements from z
isequal(x2, x1)

ans = logical
   1

x3 = z;
x3(end-4:end) = []; % I know how long y was, remove that many elements from z
isequal(x3, x1)

ans = logical
   1

Answer 2

[Jan]

pu = @(x, y) [x,y];

a = pu(1, 2:3)

a = 1×3

1 2 3

b = pu(1:2, 3)

b = 1×3

1 2 3

isequal(a, b)

ans = logical
   1

This means: No, you cannot decide, what the inputs have been based on the output.