For-loop using only last iteration
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My code is as follows:
clearvars;
zeros(1,10); %Create a row vector with 10 zeros
for i=1:1:10
A = [i.^2];
C = zeros(1,10) + A;
end
disp(C)
Now, what I originally wanted to do was add the square of each iteration on the respective i-th column of the zeros(1,10) vector, instead I get i = 10 for every position and I don't understand why it doesn't do it for each iteration but only for the last one.
3 个评论
What are you trying to do ? Add the squares of the numbers from 1 to 10 ?
x = 1:10;
s = sum(x.^2)
Or do you want
x = 1:10;
s = cumsum(x.^2)
Giuliano
2022-10-25
x = (1:10).^2
采纳的回答
It is necessary to index into ‘C’, however I do not know what you want as the result.
% clearvars;
C = zeros(1,10); %Create a row vector with 10 zeros
for i=1:1:10
A = [i.^2];
C(i) = A;
end
disp(C)
.
3 个评论
Star Strider
2022-10-25
@Giuliano — As always, my pleasure!
The problem is that you did not index ‘C’, so every iteration overwrote the previous iteration.
I also assigned the zeros call before the loop to ‘C’ since your obvious intent was to pre-allocate it. That should always be done if at all possible, so that is good programming practice there.
更多回答（1 个）
Inside the loop
C = zeros(1,10) + A;
the function "zeros(1,10)" is assigning zeros to the entire C vector, then adding A.
Also, the command before the for loop "zeros(1,10)" does not save the 10 element vector.
I think what you intended is something like:
C = zeros(1,10); %Create a row vector with 10 zeros
for i=1:1:10
A = [i.^2];
C(i) = A;
end
3 个评论
Giuliano
2022-10-25
indexing instructs the program to put the value of "A" in the 'ith" position in the C array.
So inside the for loop, when i=1, A is computed as 1^2 = 1, and put in location C(1).
On the next ittereation of the loop, i=2, A=2^2 = 4 and is stored in position 2 in the C vector (i.e. C(2) ).
When i=3, A=3^2 = 9 and is stored in C(3), etc.
So, the way you had the loop coded, C was assigned a value of zeros(1,10), or ten zeros, then it added A to the vector, which adds A to all 10 elements of C.
Each itteration of the loop assigned ten zeros to C, then added A, so only the last itteration retained any values, and they were 10 copies of A.
so in the first itteration of the loop, i=1, A=1 and C = 1 1 1 1 1 1 1 1 1 1
In the second itteration of the loop, i=2, A=4, and C = 4 4 4 4 4 4 4 4 4 4,
etc.
Giuliano
2022-10-25
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