A problem with grammar
4 次查看(过去 30 天)
显示 更早的评论
Here is a function I wrote, with N beeing a scalar (example 6) and x a vector (example [0;1;2;3;4;5]). Y1 gets input as 0 in the example and Y2 as (36-x^2)
function [ sort1 , sort2 ] = front(N,x)
Y1=input('frontière inférieure: ','s') ;
s1=vectorize(Y1);
Y2=input('frontière supérieure: ','s');
s2=vectorize(Y2);
sort1= ones(1,N+1);
for d=1:1:N;
b=x(d);
a=s1(b);
sort1(d)=a;
end
sort2= ones(1,N+1);
for d=1:1:N;
b=x(d);
a=s2(b);
sort2(d)=a;
end
end
My problem are the lines a=s1(b) and a=s2(b). My goal is to inject the values of the vector x one by one in the expressions s1 and s2, but: 1)I am not sure my semantics are correct? 2)How does this program reacts when it receives a scalar instead of an expression?
This is what I get when i run the program:
frontière inférieure: 0
frontière supérieure: 36-x^2
??? Attempted to access s1(0); index must be a positive integer
or logical.
Error in ==> front at 12
a=s1(b);
0 个评论
采纳的回答
Davide Ferraro
2012-7-26
Not sure that I've fully understood your goal. From what I've got you may try to use the EVAL function to evaluate your string. This is not the most elegant approach but can be a starting point:
function [ sort1 ] = front(N,x)
Y1=input('frontière inférieure: ','s') ;
s1=vectorize(Y1);
if ~isnan(str2double(s1))
sort1 = repmat(eval(s1),1,N+1);
else
sort1 = eval(s1);
end
end
There's margin to improve the code:
0 个评论
更多回答(4 个)
Image Analyst
2012-7-26
Yeah, a little problem with grammar - we use "syntax" when talking about that in regards to computer language rather than "grammar" or "semantics." But the problem is not a syntax problem - the real problem is that you're passing in x, which has a zero as one of its elements and you're trying to use that as an index for s1. You can't have zero as an index. So fix your x to have just positive integers, or make it a logical vector, and you should be okay.
0 个评论
Image Analyst
2012-7-26
x=[0,1,2,3,4]
so,
b=x(d);
and N = 5, that means b will equal 0, 1, 2, 3, and 4 as the loop iterates. OK, let's look at the first iteration. You say
a=s1(b);
but b = x(1) in the first iteration, which means b = 0 because x(1) = 0. So now you're saying
a = s1(0);
That's just not allowed. You can't have 0 as an index. I hope it's clear now.
0 个评论
另请参阅
产品
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!