Trying to fix the loop

조회 수: 3 (최근 30일)
Cameron Paterson
Cameron Paterson 2021년 11월 16일
편집: Jan 2021년 11월 16일
I am trying to change my code so when it loops more than five times the error message changes to Invalid vehicle type! Re-enter vehicle type (Compact, Midsize, Truck, Industrial, Military)
I have been trying to just use the variable n>5 under the for loop for this but keep getting error messages
Vehicle_Type=input('Enter the Vehicle Type','s')
for n=1:5
switch Vehicle_Type
case{'compact','Compact'}
fprintf('1\n');
break
case{'midsize','Midsize'}
fprintf('2\n');
break
case{'truck','Truck'}
fprintf('3\n');
break
case{'industrial','Industrial'}
fprintf('4\n');
break
case{'military','Military'}
fprintf('4\n');
break
otherwise
fprintf('Vehicle type is not recognized. The program stops. Rerun the program again\n')
end
end
  댓글 수: 1
Awais Saeed
Awais Saeed 2021년 11월 16일
what actually are you trying to do? Your loop does not make any sense. Your user input should be inside the loop to ask for the input for all iterations.

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Jan
Jan 2021년 11월 16일
편집: Jan 2021년 11월 16일
nFail = 0;
for n = 1:5
Vehicle_Type = input('Enter the Vehicle Type', 's');
switch lower(Vehicle_Type)
case 'compact'
fprintf('1\n');
break
case 'midsize'
fprintf('2\n');
break
case 'truck'
fprintf('3\n');
break
case 'industrial'
fprintf('4\n');
break
case 'military'
fprintf('4\n');
break
otherwise
fprintf('Vehicle type is not recognized. The program stops. Rerun the program again\n');
nFail = nFail + 1;
end
end
if nFail == 5
fprintf('bad\n');
end
I'd solve this without a for loop:
found = 0;
nFail = 0;
while ~any(found) && nFail < 5
Vehicle_Type = input('Enter the Vehicle Type', 's');
switch lower(Vehicle_Type)
case 'compact'
found = 1;
case 'midsize'
found = 2;
case 'truck'
found = 3;
case {'industrial', 'military'}
found = 4;
otherwise
fprintf('Vehicle type is not recognized. Again...\n');
nFail = nFail + 1;
end
end
if found
fprintf('%d\n', found);
else
fprintf('failed too often\n');
end

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