# In order to code HDL, how to avoid using break statements in loops

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Life is Wonderful am 11 Nov. 2022
Bearbeitet: Life is Wonderful am 11 Nov. 2022
Since I'm seeking for HDL code that doesn't allow the break, continue statement, I need help with the following code logic to stop the break statement from being executed.
case 1:
clc;
eps = 5;
x = 1:5;
y = zeros(5,1);
fprintf('%10s|%10s|\n----------+----------+\n','n','y(n)');
n| y(n)| ----------+----------+
for idx = 1:length(x)
y(idx) = x(idx) + 1;
if(y(idx) > eps)
break;
end
fprintf('%10d|%10d|\n',idx,y(idx));
end
1| 2| 2| 3| 3| 4| 4| 5|
case : 2
clc;
eps = 5;
x = 1:5;
y = zeros(5,1);
fprintf('%10s|%10s|\n----------+----------+\n','n','y(n)');
n| y(n)| ----------+----------+
for idx = 1:length(x)
y(idx) = x(idx) + 1;
if(y(idx) > eps)
idx = 1;
end
fprintf('%10d|%10d|\n',idx,y(idx));
end
1| 2| 2| 3| 3| 4| 4| 5| 1| 2|
Because more iterations are being executed, case 2's results are incorrect.
Thank you!!
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### Akzeptierte Antwort

Life is Wonderful am 11 Nov. 2022
Bearbeitet: Life is Wonderful am 11 Nov. 2022
I think i have the solution ,
x = 1:5;
eps = 5;
y = zeros(1,1);
fprintf('%10s|%10s|\n----------+----------+\n','n','y(n)');
n| y(n)| ----------+----------+
idx = 1;
while (y < eps)
y = x(idx) + 1;
fprintf('%10d|%10d|\n',idx,y);
idx = idx + 1;
end
1| 2| 2| 3| 3| 4| 4| 5|
Any fresh suggestion for improvement is welcome, and I am open to it.
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