MATLAB判断一个数是否是质数

Question

程序编好了，但是输入的不论质数合数都不能显示出来，请各位前辈指点一下，看看哪里出错了？

function [Ldiv,f]=prime(n)
% syntax: function [Ldiv,f]=prime(n)
% Check if n is prime
% Input:
% n: integer
% Output:
% Ldiv: logical; if Ldiv=true a divisor is found
% f: divisor (in case Ldiv=true)

% logical: a divisor is found
Ldiv=false;
% factor; start in while loop with value f=2
f=1;

% repeat in case no divisor is found (yet)
% AND f less than or equal to n-1
while ~Ldiv & f<=n-1
f=f+1;
if mod(n,f)==0
% a divisor is found
Ldiv=true;
end
end
% a divisor is found OR f is larger than n-1

if ~Ldiv
% no divisor is found, so n is prime
disp([num2str(n),' is prime'])
end

Answer 1

Fix
follow
f=2;%从二开始
while
~Ldiv
&
f<=n-1
if
mod(n,f)==0
%
a
divisor
is
found
Ldiv=true;
end
f=f+1;%在这里加，倘若按你原先的程序走的话，会碰到mod(n,n)==0这是一个恒成立的式子
end

Answer 2

用这么复杂？直接t=isprime(X); 如果返回值为1则为质数，为0则不是

Answer 3

Fix follow

f=2;%从二开始
while ~Ldiv & f<=n-1
if mod(n,f)==0
% a divisor is found
Ldiv=true;
end

f=f+1;%在这里加，倘若按你原先的程序走的话，会碰到mod(n,n)==0这是一个恒成立的式子
end

Answer 4

Fix
follow
f=2;%从二开始
while
~Ldiv
&
f<=n-1
if
mod(n,f)==0
%
a
divisor
is
found
Ldiv=true;
end
f=f+1;%在这里加，倘若按你原先的程序走的话，会碰到mod(n,n)==0这是一个恒成立的式子
end

Answer 5

Fix
follow
f=2;%从二开始
while
~Ldiv
&
f<=n-1
if
mod(n,f)==0
%
a
divisor
is
found
Ldiv=true;
end
f=f+1;%在这里加，倘若按你原先的程序走的话，会碰到mod(n,n)==0这是一个恒成立的式子
end