Help to find errors of if loop?

Example Complex: for,if... loop;

clear;clc;
A=[400; 900; 200; 300; 100];
k=[4;1]; % index matrix
c=[11];
e=zeros(0);
for j=0:(length(k)-1);
   b(j+1,:) = A(k(j+1,:), :);  % call vector from index
   if b(j+1,:)>200
       c=union(c,b(j+1,:));
   elseif (b(j+1,:)+100)>400
       e=union(e,b(j+1,:));
   end
end

I try to run the simpe above code to understand: if elseif loop.

Explanation of code (for ...end):

-j=0-->k(1,:)=4-->b(4,:)=A(4,:)=300 [get value from matrix A at index 4]

-j=1-->k(2,:)=1-->b(1,:)=A(1,:)=400 [get value from matrix A at index 1]

Finally we will have result matrix b=[300;400]----> GOOD

But for the (if ....elseif ...end), i hope that the result matrix e=[400], but when I run the code matrix e=[] ?????????? Can you help me where the error?

My understanding of all loops:

-j=0-->k(1,:)=4-->b(4,:)=A(4,:)=300

if b(4,:)=300>200 ---> c=[11 300]

elseif b(4,:)=300+100=400>400 : NO--->e=[]

-j=1-->k(2,:)=1-->b(1,:)=A(1,:)=400

if b(1,:)=400>200 ---> c=[11 300 400]

elseif b(1,:)=400+100=500>400: YES --->e=[400]

Finally: e=[400] : as my understanding? (How can i fix the code to get the result as my understanding)

[Jan]

Note that there are no "if loops". Only for and while are loops, while if branches according to a condition.

Inventing an own syntax like

 -j=0-->k(1,:)=4-->b(4,:)=A(4,:)=300 [get value from matrix A at index 4]

is not useful for a discussion as long, as you do not define what the symbols mean. "-j=0-->k(1.:)=4" ? It would not be smart, if I guess, what this means. Why is the valid Matlab syntax not sufficient to explain the problem? You can assume that the readers are familiar with it.

[Adam]

For starters j = 0 is not a valid index into an array and

e=zeros(0);

just creates an empty matrix.

[Guillaume]

@Adam, the j=0 is not a problem as all indexing is done with j+1. Of course, rather than going from 0 to numel(k)-1 and then adding one to all the values for indexing, it would be a lot simpler to just go from 1 to numel(k) and not add anything:

for j = 1:numel(k)
  b(j) = A(k(j, :), :);

is a lot simpler.

[ha ha]

@Guillaume. good friend. You understand well.

@Adam: --> mean: refer (or we have)

Answer 1

[Jan]

Do you know the debugger? You can set a break point in the first line and step through the code line by line. This will reveal directly, what happens inside the code.

In the second iteration b is the vector [300; 400]. Then:

if b(j+1,:) > 200

has a vector as condition. Note that if requires a scalar as argument, and therefore Matlab inserts this internally:

cond = (b(j+1,:) > 200);
if (all(cond(:)) && ~isempty(cond)

Here both elements of b are greater than 200, such that the code might do what you expect - by accident.

But the main problem remains, that you seem to assume, that the elseif branch is executed even if the if branch was already. But this is not the meaning of elseif.

I cannot guess, how you want to treat the problem of the vector input for the condition. But maybe it is enough already to replace if ... elseif ... end by if ... end, if ... end.

Note: The intention of the code is not clear. I guess boldly, that it can be simplified massively, perhaps by:

Ak = A(k);
c = unique([11; Ak(Ak > 200)]);
e = unique(Ak(Ak > 300));

[ha ha]

you are correct. just replace "if ... elseif ... end" by "if ... end, if ... end." . I will get the expected answer