Help with the output of my function
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Hi, I have written the following function which works fine when I pass it a value such as 2. But when I say something like:
>> i = -2.5:0.1:2.5;
>> p = myfun1(i);
I get there error:
Error in myfun1 (line 3)
if abs(t) <= 1
Output argument "result" (and maybe others) not assigned during call to "/Users/mbxs3/Desktop/VCU/VCU Spring 2014/EGRE 335 - Signals and Systems/MATLAB/myfun1.m>myfun1".
Here is my function code:
function result = myfun1(t)
if abs(t) <= 1
t1 = (2-abs(2*t));
result = t1;
elseif t < -1
t2 = 0;
result = t2;
elseif t > 1
t3 = 1;
result = t3;
end
采纳的回答
更多回答(3 个)
This is because when you input t as a vector, unless all of the values in vector satisfy the condition, it will not go through any of the if ... end statements.
When this happens, the value for result does not get declared and thats why you'll get error.
If you tell us, what you're trying to do in this code, maybe people can help you in fixing the issue
Wayne King
2014-2-1
编辑:Wayne King
2014-2-1
First don't use i, i the unit imaginary. That can often cause problems.
Next, you are giving it a vector input. What do you actually want to know when you write
abs(t) <=1
Do you want to know if there are ANY elements in the input that are less than 1, or whether ALL elements are?
If so use, any() and all()
if all(abs(t)<=1)
result = (2-abs(2*t));
end
or
if any(abs(t)<=1);
result = (2-abs(2*t));
end
Stephen
2014-2-1
0 个投票
2 个评论
Amit
2014-2-1
You should try Walter's method. That will give you the result.
And using i and j are choices by the user And Wayne's comment was just a reminder to you just in case you dont know.
Walter Roberson
2014-2-1
The code I gave in my answer will output the values you would like.
By default, i and j are both initialized to sqrt(-1), as variables. You can change either or both. However, people tend to confuse "i" used as a loop variable and "i" used as sqrt(-1), so it is usually clearer to avoid using "i" as a variable.
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