How can I create a 16 round for loop in this problem?

% I need different "In" on every stage and key in each stage would be ciphertext of previous stage

clear all;
clc;

Key = randi([0,1],[1,4])
In1  = randi([0,1],[1,4])
Ciphertext1 = xor(Key,In1)

Key1 = Ciphertext1
In2  = randi([0,1],[1,4])
Ciphertext2 = xor(Key1,In2)

Key2 = Ciphertext2
In3  = randi([0,1],[1,4])
Ciphertext3 = xor(Key2,In3)

Key3 = Ciphertext3
In4  = randi([0,1],[1,4])
Ciphertext4 = xor(Key3,In4)

Key4 = Ciphertext4
In5  = randi([0,1],[1,4])
Ciphertext5 = xor(Key4,In5)

Key5 = Ciphertext5
In6  = randi([0,1],[1,4])
Ciphertext6 = xor(Key5,In6)

Key6 = Ciphertext6
In7  = randi([0,1],[1,4])
Ciphertext7 = xor(Key6,In7)

Key7 = Ciphertext7
In8  = randi([0,1],[1,4])
Ciphertext8 = xor(Key7,In8)

Key8 = Ciphertext8
In9  = randi([0,1],[1,4])
Ciphertext9 = xor(Key8,In9)

Key9  = Ciphertext9
In10  = randi([0,1],[1,4])
Ciphertext10 = xor(Key9,In10)

Key10 = Ciphertext10
In11  = randi([0,1],[1,4])
Ciphertext11 = xor(Key10,In11)

Key11 = Ciphertext11
In12  = randi([0,1],[1,4])
Ciphertext12 = xor(Key11,In12)

Key12 = Ciphertext12
In13  = randi([0,1],[1,4])
Ciphertext13 = xor(Key12,In13)

Key13 = Ciphertext13
In14  = randi([0,1],[1,4])
Ciphertext14 = xor(Key13,In14)

Key14 = Ciphertext14
In15  = randi([0,1],[1,4])
Ciphertext15 = xor(Key14,In15)

Key15 = Ciphertext15
In16  = randi([0,1],[1,4])
Ciphertext = xor(Key15,In16)

Answer 1

[the cyclist]

Use a cell array to store the values:

for k = 1:16
    Key{k} = randi([0,1],[1,4]);
    In{k}  = randi([0,1],[1,4]);
    Ciphertext{k} = xor(Key{k},In{k});
end

[Md. Atiqur Rahman]

thanks for fast reply. But, I need different "In" on every stage and key in each stage would be ciphertext of previous stage