what does the expression " if ( ) && ( ) " do

Question

ad lyn 2021-10-29

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1574548-what-does-the-expression-if-do

评论： ad lyn 2021-10-29

 if (i>=njmp1) && (i<=njmp2) 
                        fcs(i)=a*i+b;
                    elseif (i>njmp2)
                        fcs(i)=fc2;
                        
                    end

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

James Tursa 2021-10-29

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1574548-what-does-the-expression-if-do#answer_819838

&& is the "logical and" operator. You can read about it here:

https://www.mathworks.com/help/matlab/ref/logicaloperatorsshortcircuit.html

1 个评论
显示 -1更早的评论隐藏 -1更早的评论

ad lyn 2021-10-29

but when we use it in the previous statement is it gonna keep the "AND" fonctionnality?

请先登录，再进行评论。

Answer 2

Walter Roberson 2021-10-29

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1574548-what-does-the-expression-if-do#answer_819898

在 MATLAB Online 中打开

Except for line numbering and issues about when interruptions are permitted, the code you posted is equivalent to

 if (i>=njmp1)
     if (i<=njmp2) 
        fcs(i)=a*i+b;
     elseif (i>njmp2)
        fcs(i)=fc2;
     else
         %do nothing
     end
 elseif (i>njmp2)
    fcs(i)=fc2;
 else
     %do nothing
 end

complete with the implication that i<=njmp2 will not be executed at all if i>=njmp1 is false.

Imagine that you have a situation where you have a variable in which a non-nan second element is to be used, provided that a second element exists:

 ub = inf;
 if length(bounds) > 1
    if ~isnan(bounds(2))
      ub = bounds(2);
    end
 end

this cannot be coded as

 ub = inf;
 if length(bounds) > 1 & ~isnan(bounds(2))
    ub = bounds(2);
 end

because the & operator always evaluates all if its parts, and so would always try to access bounds(2) even if length(bounds)>1 were false. The & operator is not smart about stopping if the condition cannot be satisfied. But

 ub = inf;
 if length(bounds) > 1 && ~isbounds(2))
   ub = bounds(2);
 end

is fine. The && operator is smart about not evaluating the second operand if the first is false -- not just "smart" but guarantees that the second part will not be evaluated. (In such a situation to just say it was "smart" implies that it is a matter of optimization, that it thinks it is "more efficient" not to evaluate the second operand -- but that implies that in some circumstances, perhaps involving parallel processing, if it happened to be more efficient to evaluate the second clause, that it could potentially be evaluated. But the && operand promises that the second operand will not be evaluated if the first one is false, so it is safe to use && with chains of tests that rule out the possibility of exceptional behaviour.)