code is too long

1 次查看（过去 30 天）
Nicolò
Nicolò 2013-3-31
a = 4;
b = input ('2+2=? ')
if b == a
'ok'
else
b = input ('retry ')
if b == a
'ok'
else
b = input ('retry ')
if b == a
'ok'
else 'NOK'
end
end
end

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采纳的回答

Image Analyst
Image Analyst 2013-3-31
编辑：Image Analyst 2013-3-31
I think you want
a = 4;
b = 0;
while b ~= a
b = input ('2+2=? ');
if b == a
break;
end
fprintf('Incorrect. Try again.\n');
end
fprintf('Correct.\n');
  1 个评论
Image Analyst
Image Analyst 2013-3-31
If you want to limit it to three tries, add a counter:
a = 4;
b = 0;
counter = 1;
while b ~= a && counter <= 3
b = input ('2+2=? ');
if b == a
break;
end
fprintf('Incorrect. Try again.\n');
counter = counter + 1;
end
if counter <= 3
fprintf('Correct.\n');
end

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更多回答（1 个）

Nicolò
Nicolò 2013-3-31
Yes, but I would like to repeat this code up to three times if the number entered is incorrect
  2 个评论
Image Analyst
Image Analyst 2013-3-31
This is not an answer to your question. It should have been a comment to my answer so I'll put the response there.
Nicolò
Nicolò 2013-3-31
thanks

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