How come this codes works, but the other one doesn't.

1 次查看(过去 30 天)
(WORKS) %% variables
V = [17.43 11.46 8.11 4.23 2.29 30.18 19.84 14.05 7.34 3.96 34.85 22.91 16.22 8.47 4.57 13.18 19.84 14.05 7.34 3.96 17.43 11.46 8.11 4.23 2.29];
val = input('Choose a distance between 0 and 36 feet \n');
tmp = abs(V-val);
[xp xp] = min(tmp);
C = V(xp);
%% conditional
if xp <= 25 && xp >=21;
A = 75;
elseif xp <= 20 && xp >=16;
A = 60;
elseif xp <= 15 && xp >=11;
A = 45;
elseif xp <= 10 && xp >=6;
A = 30;
elseif xp <= 5 && xp >=1;
A = 15;
end
if xp == 1|| xp == 6 xp == 11 xp == 16 xp == 21;
M = 1;
elseif xp == 2 xp == 7 xp == 12|| xp == 17|| xp == 22;
M=2;
elseif xp == 3 || xp == 8 || xp == 13 || xp == 18 || xp == 23;
M=3;
elseif xp == 4 xp == 9 xp == 14 xp == 19 xp == 24;
M=4;
elseif xp == 5 || xp == 10 || xp == 15 || xp == 20 || xp == 25;
M =5;
end
if val== 0; M = 1; A = 0; end
%% display
fprintf('The setting that best fits your distance is %0.0f degrees and the %0.0fº string position, which wil give you %0.2f feet', A, M, C);
(DOES NOT WORK)(BELOW)
V = [13.36 12.37 11.36 12.40 10.30 11.43 7.70 9.19 4.60 6.66];
val = input('Choose a distance between 0 and 36 feet \n');
tmp = abs(V-val);
[xp xp] = min(tmp);
C = V(xp);
%% conditional
if xp <= 10 && xp >=9;
A = 75;
elseif xp <=8 && xp >=7;
A = 60;
elseif xp <= 6 && xp >=5;
A = 45;
elseif xp <= 4 && xp >=3;
A = 30;
elseif xp <= 2 && xp >=1;
A = 15;
end
if xp == 1|| xp == 6
M=1;
elseif xp == 2 xp == 7;
M=2;
elseif xp == 3 xp == 8;
M=3;
elseif xp == 4 xp == 9;
M=4;
elseif xp == 5 xp == 10;
M=5;
end
if val== 0; M = 1; A = 0; end
%% display
fprintf('The setting that best fits your distance is %0.0f degrees and the %0.0fº string position, which wil give you %0.2f feet', A, M, C);

回答(1 个)

Image Analyst
Image Analyst 2016-5-5
I've never seen %0.0f and don't know what you're thinking. Just use %f or something without zeros in it. If you want integers, use %d.
  1 个评论
Stephen23
Stephen23 2016-5-5
编辑:Stephen23 2016-5-5
I don't know how '%0.0f' could be useful, but '%.0f' can be useful to keep the length of the output correct, e.g. when it should be equal to the number of digits:
>> sprintf('%d',1e19)
ans =
1.000000e+19
>> sprintf('%.0f',1e19)
ans =
10000000000000000000
This is sometimes a more reliable way to count digits than log10 or the like.

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