MATLAB initializing variables in next for loop

Question

x=[3 1 1 -5 -2 0 1 2 -2 2];
A=[4 2 6; 0 1 -3; -2 5 -2];
B=[-2 3 2; 1 5 5; -3 1 0];

sum=0;
for i=2:3
    sum_j=0;
    for j=1:2
        sum_j=sum_j+A(1,j)*B(j,i);
    end
    sum=sum+A(2,i)*sum_j;
end
fprintf('(c) %g\n',sum);

>> (c) -32

-32 is a correct answer. However, if I initialize sum_j=0 outside of the the loop, it returns a different value.

sum=0;
sum_j=0;
for i=2:3
    for j=1:2
        sum_j=sum_j+A(1,j)*B(j,i);
    end
    sum=sum+A(2,i)*sum_j;
end
fprintf('(c) %g\n',sum);

>> (c) -98

Can anyone explain why this is happening?

This is obviously what is happening. In the first place you set sum_j to zero for every "i" and in the second place you don't. — Konstantinos Monachopoulos
– Konstantinos Monachopoulos, Commented Mar 5, 2017 at 12:27

lkdhruw · Accepted Answer · 2017-03-05 07:07:49Z

1

In your first code sum_j gets reinitialized in the loop for i and in second code sum_j carry the value in each loop. Here is a simulation of your codes

first code:

sum = 0
i=2:
    sum_j = 0
i=2,j=1
    sum_j = 0 + 4*3 = 12
i=2,j=2
    sum_j = 12 + 2*5 = 22
  sum = 0 + 1*22 = 22
i=3
    sum_j = 0
i=3,j=1
    sum_j = 0 + 4*2 = 8
i=3,j=2
    sum_j = 8 + 2*5 = 18
  sum = 22 + -3*18 = -32

Second code

sum = 0
sum_j = 0
i=2:
i=2,j=1
    sum_j = 0 + 4*3 = 12
i=2,j=2
    sum_j = 12 + 2*5 = 22
  sum = 0 + 1*22 = 22
i=3
i=3,j=1
    sum_j = 22 + 4*2 = 30
i=3,j=2
    sum_j = 30 + 2*5 = 40
  sum = 22 + -3*40 = -98

edited Mar 5, 2017 at 7:07

answered Mar 5, 2017 at 6:42

lkdhruw

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1 Comment

Eric Kim Over a year ago

Great!! that perfectly explains

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