is:Conversion to logical form function_handle is not possible.
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Hi I want to do a comparison with function handle. For example : Fun2=@(x) (4*x+1) While @(x)Fun2(x) >3 do something
The error is:Conversion to logical form function_handle is not possible. Thank a lot
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Walter Roberson
2020-5-10
while Fun2(x) > 3
and be sure to update x inside the loop
2 个评论
Tara Behmanesh
2020-5-10
Walter Roberson
2020-5-10
编辑：Walter Roberson
2020-5-10
The message is correct. You do not define a or b before you try to use them in
while fun2(e,a,b)>10^(-6)||fun2(e,a,b)<-10^(-6)
Reminder: when you use
fun1=@(a,b)(b-a)./2;
then that does not define variables a or b: it defines only fun1 and it defines fun1 as being an anonymous function that accepts two positional parameters that for convenience are referred to as a and b.
fun1=@(a,b)(b-a)./2;
is coded sort of like
struct( 'formula', #(_Parameter{2}-_Parameter{1})./2)# , 'ParameterNames', {'a', 'b'})
in the sense that the execution of the code depends only on parameter position and the names of the parameters are (nearly) only used for display purposes.
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