Dealing with wrong input

How do I avoid this error? I would like for users, if they enter a string instead of number to told "Those are not the instructions, please try again"

[Nicolas B.]

Can you provide your example code?

[Lars Morsink]

[Rik]

Please post your code as code, not as a screenshot. You can find an explanation of the layout tools here.

[Lars Morsink]

clc, 
clear
Secret_Number = randi(100);
Attempts = 0;
Max_Attempts = 10;
Guess = -1; % The variable needs to be established to avoid an error in the 'while...end' loop and -1 guarantees that it is not equal to the secret number
disp (' ')
disp ('Instructions: Guess a number between 1 and 100. You have 10 guesses and a hint after each guess. Good luck!')

while (Attempts < Max_Attempts) && (Secret_Number ~= Guess)
    disp (' ')
    Guess = input('Please enter your guess: ');
    Attempts = Attempts + 1;
    
    if Secret_Number == Guess
    disp (' ')
    disp ('Congratulations! You guessed the secret number!')
    Attempts
    elseif (Guess > Secret_Number) && (Attempts < Max_Attempts);
        disp (' ')
        disp ('Hint: Guess lower') 
    else
        disp (' ')
        disp ('Hint: Guess higher')
    end
end  

if Attempts == Max_Attempts & (Secret_Number ~= Guess);
    disp (' ')
    disp ('Game Over! You ran out of guesses. Please play again.')
    Secret_Number
end

Answer 1

[Stephen23]

str = '';
num = NaN;
while isnan(num)
    num = str2double(input([str,'Enter a number: '],'s'));
    str = 'Please try again! ';
end

And tested:

Enter a number: hello
Please try again! Enter a number: world
Please try again! Enter a number: 42

Answer 2

[Nicolas B.]

Okay, then I would use the exception catching method:

str = '';
num = NaN;
while isnan(num)
    try
        num = input('Please enter your guess: ');
    catch
        disp('You must give a valid number!');
        num = NaN;
    end
end

[Rik]

That's also possible, although I think it would make slightly more sense to use Stephen's method, because then it also isn't possible to provide a variable name.

If you define a=9; before your loop, you can enter a in the prompt and it will pass the test. With Stephen's code, that is not possible.