Why does my if statement do the opposite?

조회 수: 4 (최근 30일)
Erwin Avendaño
Erwin Avendaño 2017년 11월 8일
댓글: David Goodmanson 2017년 11월 8일
My if statement is asking that if h is negative, you must do that operation if not the other but do the opposite, do not I understand?
t=b4
e=b3
f=b2
h=e^2-4*t*f
if h==-h
qc=3
p=(h)*(-1)
x5=(sqrt(h))/(2*t);
x6=(-e/(2*t));
x7=(-sqrt(h))/(2*t);
x8=(-h/(2*t));
else
qd=4
x5=(-e+sqrt(h))/(2*t);
x6=(-e-sqrt(h))/(2*t);
end

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David Goodmanson
David Goodmanson 2017년 11월 8일
Hi Erwin
The test h==-h is not a test for whether h is negative. This test is the same as 2*h==0 and always fails unless h = 0. Take a look at h<0, or h<=0 if you intend the condition to work for h=0 as well as negative h.
  댓글 수: 3
Erwin Avendaño
Erwin Avendaño 2017년 11월 8일
Is these two conditions okay? if (qb==2 && qd==4)
David Goodmanson
David Goodmanson 2017년 11월 8일
They are the right syntax anyway, assuming that qb and qd are scalars. I tend to just use & in every situation because it always works and I don't think it has slowed things down for what I ordinarily do.

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