Help with quadriatic formula?
显示 更早的评论
Hi everyone,
We are to create a function using the editor to solve a quadratic formula. The hard part is the user is to be able to input just the equation ie: Ax^2+Bx+C=0 (not the values of ABC). (Below is what I have so far)
____***_*%The user input the formula they want solved_____
function [xpos, xneg] = Roots(equation)
%Quadratic equation format is = Ax^2 + Bx + C = 0
A = strtok(equation, 'x');
B =
C =
%This is the discriminant
D = B .^2 - 4 .* A .* C;
%These are the equation to find the positive and negative values of x
xpos = (-B + sqrt(D)) ./ (2 .* A)
xneg = (-B - sqrt(D)) ./ (2 .* A)
end
3 个评论
Matt Kindig
2013-9-12
To clarify, your input is something like this following, correct?
equation = '3x^2 + 5x + 12=0'
Daniel Shub
2013-9-17
There are lots of parsers for quadratic equations based on regular expressions. This might be easier than using strtok
dpb
2013-9-17
Oh, indeed! The OPs instructor is, however, apparently not amenable to alternate solutions than those suggested based on his comments later on.
采纳的回答
dpb
2013-9-12
You showed enough I'll do more than normally would on first response...
s=inputdlg('Enter quadratic: ','Quadratic Solver Input',1,{'ax^2+bx+c'});
8 个评论
MAB
2013-9-16
dpb
2013-9-16
How carefully are you controlling the format of the input the user provides? If the structure of the equation can be assumed to match a given format it's possible to simply do something like
>> s='3x^2+5x+1';
>> c=sscanf(s,'%f x^2 + %f x + %f')'
c =
3 5 1
>>
If you need to or want to use strtok() to parse substrings, simply use the optional "remainder" output and a while loop to return subsequent tokens until it becomes the empty string.
[t,r]=strtok(s);
% process token appropriately here
while ~isempty(r)
[t,r]=strtok(s);
% process the next token appropriately here
end
dpb
2013-9-17
That's just evil imo... :)
I don't see any good way to actually parse any of the other numeric values after the first with strtok() only--it just doesn't parse regularly based on the rules strtok() uses.
You could do a two-step process of first separating the terms based on defining the delimiter string as '+=' as the first pass then delimiter "x" on that token for a and b. But c is still an oddball that then parses on the first pass owing to using += for the primary delimiter.
Sotoo,
delim='+=';
[a,r]=strtok(s,delim); a=strtok(a,'x');
[b,r]=strtok(r,delim); b=strtok(b,'x');
c=strtok(r,delim);
Of course, all are still strings, not numeric; you still need a str2num() around each or sscanf() or somesuch to get the actual numeric value.
All in all, I'm not impressed by the instructions received...
dpb
2013-9-17
Well, it'll be interesting to hear if the instructor actually can come up with a realistic parser under the given rules himself...keep us posted.
MAB
2013-10-17
dpb
2013-10-17
The form doesn't fit the required input of an explicit constant.
Only way to deal with it will be to special-case it.
更多回答（0 个）
类别
在 帮助中心 和 File Exchange 中查找有关 Historical Contests 的更多信息
Translated by 
