if statement not displaying

%%CODE FOR INGRIEDENTS CODE
disp('Ingrediants ID#');
disp('eggs: 10');
disp('avocado: 21');
disp('bread slices: 32');
disp('peanut butter: 43');
disp(' cup of spinich: 54');
disp('1/4 tbs of butter: 65');
disp('slices of cheese: 76');
n = input('how many ingredients do you have? ');
%%FOOD INPUT
for k = 1:n
a(k) = input(sprintf('Type code of Ingrediant #%d: ',k ));
A(k) = input(sprintf('Type quantity of Ingrediant #%d: ',k ));
end
if (n) >= (3)
elseif (a(k)) == (10)
elseif (a(k)) == (21)
elseif (a(k)) ==(32)
    eggsandwich =  ((a(1)*A(1))+(a(2)*A(2))+(a(3)*A(3)))
    if eggsandwich >= (95)
            disp('egg avocado sandwich')
   end
end

is this case input is:

a(1) = 10;
A(1) = 1;
a(2)=21;
A(2)= 1;
a(3) = 32;
A(3) = 2;

eggsandwich = 95

trying to display 'egg avocado sandwich'

Answer 1

[Walter Roberson]

if (n) >= (3)

You are entering 3 ingredients, so n >= 3 is true, so the body of the "if" will be executed.

elseif (a(k)) == (10)

Tbe body of your "if" is empty. You will only go on to the "elseif" if the first "if" is false, but it is true. You can only get to the egggsandwich assignment statement if n >= 3 and a(k) == 10 and a(k) == 21 are all false, but a(k) == 32.

Note that the value of k is "left over" from the last value it was assigned in the for k = 1:n loop. You are not executing the if/elseif tree for every k value, only for the k that exists after the for k loop completely finishes, which in this case will be k == n.

I would suggest to you that you would be more interested in

    any(a == 10) && any(a == 21) && any(a == 32)

[Jarren Berdal]

now im getting 'Operands to the || and && operators must be convertible to logical scalar values.' error

[Walter Roberson]

If you are using the any() like I show, then that error would only occur if your a had more than one dimension.