Why didn't this 'try.. catch' work?
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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
採用された回答
その他の回答 (1 件)
Jos (10584)
2016 年 6 月 8 日
1 投票
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 件のコメント
SP Lee
2016 年 6 月 8 日
SP Lee
2016 年 6 月 8 日
Guillaume
2016 年 6 月 8 日
"MATLAB displays the relevant error message" that's all it does, display. It does not throw an error, therefore there is no error to catch.
SP Lee
2016 年 6 月 8 日
Walter Roberson
2016 年 6 月 8 日
So it caught the error and handled it. You are not going to have any success overriding that "catch".
Guillaume
2016 年 6 月 8 日
@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
2016 年 6 月 8 日
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