why if condition is not working in this code.

% code
dbstop if error
x=input('enter no1: ');
z=input('enter no2: ');
if  (x<z)
  disp('invalid input')
else
  disp('ok')
  % n=x:z;
  % for i=1:n
  %     n;
  % end
  % n
end

[dpb]

>> x=input('enter no1: ');
z=input('enter no2: ');
if  (x<z)
  disp('invalid input')
else
  disp('ok')
  % n=x:z;
  % for i=1:n
  %     n;
  % end
  % n
end
enter no1: 3
enter no2: 2
ok
>> x=input('enter no1: ');
z=input('enter no2: ');
if  (x<z)
  disp('invalid input')
else
  disp('ok')
  % n=x:z;
  % for i=1:n
  %     n;
  % end
  % n
end
enter no1: 2
enter no2: 4
invalid input
>>

Seems to work fine; what's the specific issue?

The commented-out for loop probably isn't constructed with what you intend, but there's not sufficient information provided to know what the intent is.

You may want to explore what the expression

1:n

returns for various combinations of the inputs.

[Robiul Ferdous]

in the if condition when x<z it is not showing the desired output written in the code

[Robiul Ferdous]

this is my output,but it is supposed to show 'invalid input'

 enter no1: 5
enter no2: 1
ok

n =

   Empty matrix: 1-by-0

[dpb]

It showed exactly what you coded it to show for the given inputs--

enter no1: 5
enter no2: 1

Your code again:

x=input('enter no1: ');
z=input('enter no2: ');
if  (x<z)
  disp('invalid input')
else
  disp('ok')
  ...

x <-- No1 and z <-- No2 and 5 is not less than 1 so the else block is going to be executed and that results in the 'ok' message and n never being defined.

Mayhaps if you used variable names that represent the prompts or prompts that reflect the variable names used it would be simpler to keep straight "who's who in the zoo", but the problem isn't the if (or at least a ML problem with it; you may have inadvertently written the sense of the test backwards from what you intended or somesuch).

Answer 1

[dpb]

As noted, perhaps the following would help...

>>  x=input('enter no1: ');
z=input('enter no2: ');
if  (x<z)
  fprintf('\nInvalid input:\nNo 1 (x=%g) is less than No 2 (z=%g)\n',x,z)
else
  disp('ok')
end
enter no1: 2
enter no2: 5

Invalid input:
No 1 (x=2) is less than No 2 (z=5)
>>