Where is the problem in this if-else structure?
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This is an if-else structure of a larger problem. I have simplified it down to the numerals in work here, but can't figure out where lies the problem.
when I evaluate this simple if structure, the program effortlessly outputs a value of "w=1", (The program enters the structure.)
if (1 > 0)&(3 <= 4)
w = 1;
end
But when I evaluate the same above structure inside an if, else if combination the program won't enter in the "elseif structure" for some reason, and only outputs "m=0", whereas I should also get "w=1". What am I doing wrong?
if (4 <= 4)& (3 <= 4)
m= 0;
elseif (1 > 0)&(3 <= 4)
w = 1;
end
Matlab always evaluates:
(4 <= 4)& (3 <= 4)
equals to logical 1. Why won't it then enter inside the if, else-if structure?
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回答 (2 件)
per isakson
2016 年 8 月 28 日
編集済み: per isakson
2016 年 8 月 30 日
"whereas I should also get "w=1"."   No it should not, because doc says "The statements execute only if previous expressions in the if...end block are false.". In your case a previous expression, (4 <= 4)& (3 <= 4), is true. Only the statements following the first expressions that is true will be executed.
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John BG
2016 年 8 月 29 日
編集済み: John BG
2016 年 8 月 29 日
because in your
if cond1
elseif cond2
end
cond1 and cond2 are nested, cond2 only happens if cond1 is false. Because in your example cond1 is 1, true, the if exits. It's the way it works.
Try instead
if cond1
..
end;
if cond2
..
end;
or try a switch
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John BG
3 件のコメント
John BG
2016 年 8 月 30 日
編集済み: John BG
2016 年 8 月 30 日
I didn't mention the command switch to do what he expected elseif to do but it doesn't.
I am just encouraging to try a different command, that does the same, hoping for the understanding of the basic point that 'else' implies exclusion, not the expected inclusion that Abdul wonders about.
Guillame, next comment please add something we do not know, cool?
Walter Roberson
2016 年 8 月 31 日
Guillame was mentioning something that the original poster probably did not know.
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