Why didn't this 'try.. catch' work?

17 visualizzazioni (ultimi 30 giorni)
SP Lee
SP Lee il 8 Giu 2016
Commentato: SP Lee il 8 Giu 2016
If at the prompt I enter anything other than a numerical value or previously defined variable, the 'try.. catch' block does not throw an exception. Why?
try
user_input=input('Please enter a NUMERICAL input:..\n');
catch ME1
ME1
end
disp('Life goes on..');
Instead it just keep prompting me to enter an input, until I enter a valid one

Risposta accettata

Walter Roberson
Walter Roberson il 8 Giu 2016
Read the input as a string and validate it yourself, and then you will have control over the behavior if a non-numeric value is entered.

Più risposte (1)

Jos (10584)
Jos (10584) il 8 Giu 2016
You should read the documentation of input which says:
" If the user enters an invalid expression at the prompt, then MATLAB displays the relevant error message, and then redisplays the prompt."
  7 Commenti
Guillaume
Guillaume il 8 Giu 2016
@SP Lee, from the point of view of the calling function there is no exception thrown by input and thus no exception to catch. Most likely, the way input does this is that it has its onw try_catch hence why MException.last gets changed. The calling code will never see this exception.
The fact that MException.last gets changed is a leakage of the implementation details of input, not something you should rely on. Unfortunately, there's plenty of such leakages in matlab.
Do what Walter told you to do, grab the input as a string and do your own validation:
user_input=input('Please enter a NUMERICAL input:..\n', 's'); %s to get a string
try
eval(sprintf('user_input = %s', user_input));
catch ME1
ME1
end
disp('Life goes on..');
SP Lee
SP Lee il 8 Giu 2016
Agree... Thanks all!

Accedi per commentare.

Categorie

Scopri di più su Structures in Help Center e File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by