can any one tell how this break function works in this loop??
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Abhishek sadasivan
el 15 de Sept. de 2014
Comentada: Abhishek sadasivan
el 16 de Sept. de 2014
can any one tell me how the break functions if there are two if conditions in a for loop ...
if true
% code
end
locs_Qwave=[ 100 150 90 1175 1 50]';
locs_Rwave=[116 170]';
q=0
k=1
for j=k:size(locs_Rwave)
for i=1:numel(locs_Qwave)
if (i== numel(locs_Qwave))
q=[q locs_Qwave(i)];
break;
end
if( locs_Qwave(i)>locs_Rwave(j))
q=[q locs_Qwave(i-1)];
break;
end
end
end
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Adam
el 15 de Sept. de 2014
Editada: Adam
el 15 de Sept. de 2014
The second if statement is hit at j = 1, i = 2 so locs_Qwave(1) gets appended to q and it breaks out of the loop before the first if statement kicks in at all. Then we get j = 2 and now the second if statement kicks in at i = 4, adding locs_Qwave(3) to q. Again the first condition plays no part because it is not matched before the second condition matches and breaks out of the loop.
The first of condition appears to only be checking what would be the end of the for loop anyway so it will never activate before the final iteration. All it would do on its own is add a value to q after the for loop has ended which you could do without even having a for loop.
So with the other if statement in there too the first clause will only activate if the second clause is never matched.
3 comentarios
Adam
el 16 de Sept. de 2014
You never reach the final iteration. j only runs from 1 to 2 so there are only two outer loops, the first breaks at i = 2, the second at i = 4 and that is the end of the for loop. i = 6 is never reached.
You can just add
q=[q locs_Qwave(end)];
after your for loop if you want that to always happen though. Or add locs_Qwave(end) twice if you want it to happen for each j.
Más respuestas (1)
Image Analyst
el 15 de Sept. de 2014
It looks like it should function. What's the problem? If either of those two conditions is met, it breaks out of the "i" for loop and continues on with the "j" for loop.
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