Does someone know how to print a square onto the command window using for loops????
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for 1:5
fprintf('*') % am stuck right there
Risposte (3)
Daniel Shub
il 17 Set 2013
Modificato: Daniel Shub
il 17 Set 2013
I am not sure why you want to use a loop or fprintf. The simplest, but potentially not square enough solution for a filled square
N = 5;
x = repmat(char(42), N, N);
disp(x);
*****
*****
*****
*****
*****
For an unfilled square you can do
N = 5;
x = repmat(char(42), N, N);
x(2:end-1, 2:end-1) = char(32);
disp(x);
*****
* *
* *
* *
*****
On my system I get a squarer square with
x = repmat([char(42), char(32)], N, N)
* * * * *
* * * * *
* * * * *
* * * * *
* * * * *
Depending on how perfect you want, maybe you could use a thin space or a wide space.
For a filled diamond
a = [fliplr(tril(true(N))), tril(true(N)); triu(true(N)), fliplr(triu(true(N)))];
x = repmat(char(32), 2*N, 2*N);
x(a) = char(42);
disp(x)
**
****
******
********
**********
**********
********
******
****
**
For an empty diamond
a = [fliplr(triu(tril(true(N)))), triu(tril(true(N))); tril(triu(true(N))), fliplr(tril(triu(true(N))))];
x = repmat(char(32), 2*N, 2*N);
x(a) = char(42);
disp(x)
**
* *
* *
* *
* *
* *
* *
* *
* *
**
If you really need to use fprintf
y = mat2cell(x(:), ones(numel(x), 1), 1);
fprintf([repmat('%s', 1, 2*N), '\n'], y{:})
4 Commenti
Kevin Junior
il 17 Set 2013
Daniel Shub
il 17 Set 2013
See my edit for diamonds.
Usama Tanveer
il 17 Ott 2022
What if we tried to plot this shape
Rik
il 17 Ott 2022
@Usama Tanveer The answer to this question can be found in the documentation of the plot function, which is probably the first place you should have looked. Did you?
Simon
il 17 Set 2013
0 voti
Hi!
You should start reading here: http://www.mathworks.com/help/matlab/control-flow.html This explains how to use loops and their syntax.
What do you mean with "square"?
2 Commenti
Kevin Junior
il 17 Set 2013
Walter Roberson
il 17 Set 2013
Filled or outline only?
Some of the shapes will not require nested for loops.
Kambiz Hosseinpanahi
il 26 Giu 2018
Modificato: Walter Roberson
il 26 Giu 2018
clc
clear
n=10;
A=repmat(char(42),n,n);
A(1:end-1,2:end)=char(32);
for i=2:10
A(i,i)=char(42);
end
disp(A);
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