How do you use logical operators for this problem in multiple unnested if statements?

Question

Caroline F 2022-2-26

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https://ww2.mathworks.cn/matlabcentral/answers/1659260-how-do-you-use-logical-operators-for-this-problem-in-multiple-unnested-if-statements

评论： Steven Lord 2022-2-27

在 MATLAB Online 中打开

I'm given the following:

x=0:3:21;

y=10;

z=15;

When x < y and z, the code has to dispaly "Value is less that y and z". When y<x<z, the code has to display "Value is between y and z". When x > y and z, the code has to display " Value is greater than y and z." I was able to do this with a for loop and nested if statements, but I also have to write the code using unnested if statements. My code either gives logical array instead of a displayed answer or an error statement. This is what I have so far, and it gives a logical array.

x=0:3:21;
y=10;
z=15; 
if x<y %& x<z %y is less than z, so if x is less than y, then x must also be less than z
    disp('Value is less than y and z')
    
elseif x>z & x>y %z is greater than y, so if x is greater than z, then x must also be greater than y
    disp('Value is more than y and z')
    
else y<x<z
    disp('Value is between y and z')
end
ans = 1×8 logical array
   1   1   1   1   1   1   1   1
Value is between y and z

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Answer 1

Image Analyst 2022-2-26

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1659260-how-do-you-use-logical-operators-for-this-problem-in-multiple-unnested-if-statements#answer_905435

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Replace

else y<x<z

with

elseif (y < x) && (x < z)

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Steven Lord 2022-2-27

在 MATLAB Online 中打开

Because x is a vector that won't work. You could wrap the conditions in all or any to get it to work or you could use the non-short-circuiting and operator &.

x = 1:10;
y = 0;
z = 11;
all(y < x) && all(x < z) % true
ans = logical
   1
(y < x) & (x < z) % vector of true values
ans = 1×10 logical array
   1   1   1   1   1   1   1   1   1   1
(y < x) && (x < z) % error
Operands to the logical and (&&) and or (||) operators must be convertible to logical scalar values.