I have an assignment to write a macro in LibreOffice that will code a message using XOR operation. I get the problem: Function not allowed within a procedure. when I try to run the macro.
This is the code:
REM ***** BASIC *****
Sub Main
end Sub
function izracunajHash(geslo, zacetni_hash)
zacetni_hash = 17520
hash = zacetni_hash
mask = &H00FFFFFF
dolzina = len(geslo)
If dolzina > 0 Then
for f=1 to dolzina step +1
podniz = mid(geslo,dolzina,1)
char = Asc(podniz)
hash = 33*hash + char
hash = hash AND mask
dolzina = dolzina +1
hash = hash AND &H00FFFFFF
next f
izracunajHash = hash
End function
function kodiraj(niz) //this is where the problem occurs according to LibreOffice
y = 1
if Len(niz) > 0 Then
x = Len(niz)
Do While y > (x+1)
sign = Mid(niz, y, 1)
z1 = Asc(sign)
if z1 > 31 Then
z2 = (CInt(rnd()*31))
z1 = z1 XOR z2
z1 = Chr(z1)
Mid(niz,y,1,z1)
End If
y = y + 1
Loop
End If
kodiraj = niz
End function
Thanks in advance. (I cut some unimportant code that I didn't have to write myself)
In your first function you have an if statement without a corresponding End If.
Related
I implemented merge sort in scilab with the following code:
function x = mergesortre (x)
n = length (x);
if ( n > 1 ) then
m = floor (n/2);
p = n-m;
x1 = mergesortre ( x(1:m) );
x2 = mergesortre ( x(m+1:n) );
x = merge ( x1 , x2 );
end
endfunction
function [x] = merge ( x1 , x2 )
n1 = length (x1);
n2 = length (x2);
n = n1 + n2;
x = [];
i = 1
j = 1
k = 1
while(j<=n1 && k<=n2)
if x1(j)>=x2(k)
x(i)=x2(k);
k=k+1;
i=i+1;
elseif x1(j)<x2(k)
x(i)=x1(j);
j=j+1;
i=i+1;
end
end
if (j > n1) then
x(i+1:n) = x2(k:n2);
else
x(i+1:n) = x1(j:n1);
end
endfunction
a=[5,4,3,2,1];
x=mergesortre(a);
disp x;
However in when i try to see the sorted array using the terminal window its showing only the first element,for example if my array is [5,4,3,2,1] its only giving output as 1. I need help understanding what i did wrong.
Your error is in the merge function. As i is already incremented at the end of the loop you have to concatenate starting at i:
if (j > n1) then
x(i:n) = x2(k:n2);
else
x(i:n) = x1(j:n1);
end
A recursive algorithm needs a termination condition. There is no such condition in your function mergesortre. It's likely because the declaration of your function does not used the input argument x. Try to change it like this:
function [a] = mergesortre ( a )
I guess you wanted to use a as an input and output argument.
One of the methods to compute sqrt(a), a>0 is
X(n+1) = (a + (X(n)*X(n-1))/(X(n)+X(n-1)), n = 1, 2, …,
with X0=1 and X1=a (That is, it is known that lim n-> infin of Xn = sqrt(a)
Write a function [sqa, nitr] = mySqrt(a) which implements this calculation. The function should use a while-loop, terminate when the difference between Xn+1 and Xn becomes smaller than eps(10*a), and output Xn+1 in sqa and the value of n at which the while-loop was terminated in nitr. Test your function for a = 103041.
I have written this but it does not work
function [sqa, nitr] = mySqrt (a)
%[sqa, nitr] = mySqrt (a)
% computes square root of a
% sqa = a;
sqaprev = a;
nitr = 0;
X(n+1) = (a + (X(n)*X(n-1))/(X(n)+X(n-1))); %find the second term
sqa= X(n+1)
while abs (sqaprev-sqa) >= eps (10*a)
sqaprev = sqa;
sqa = (1/2) *(sqaprev+ (a/sqaprev));
nitr = nitr + 1;
end %while
end
i get the error:
Unrecognized function or variable 'X'.
Error in mySqrt (line 7)
X(n+1) = (a + (X(n)*X(n-1))/(X(n)+X(n-1))); %find the second term
Could someone help me ?
You should start with declaring your variables and assigning them values
X(1)=1;
X(2)=a;
n=2;
Then in the loop you apply the given recursion formula, not the Heron/Babylonian formula that got from somewhere else.
According to the algorithm you presented for the square root, you can try the code below
function [sqa, n] = mySqrt(a)
n = 2; % start from 2
X = [1,a]; % initial value for sequence X
while true % repeat procedure until termination condition is reached
X(n+1) = (a + (X(n)*X(n-1)))/(X(n)+X(n-1)); % update sequence X by assigning new values
if abs(X(n+1)-X(n)) <= eps(10*a) % termination condition
break;
end
n = n + 1;
end
sqa = X(end); % return the last element of X as the square root
end
such that
>> [sqa,n] = mySqrt(a)
sqa = 321.00
n = 20
I am attempting to write a function called expSeries which uses another function factFunc to evaluate e^x. I have already written the function factFunc, as shown below:
function fact = factFunc(n)
f = 1;
for a = 1:b
f = f*a;
end
fact = f;
end
I am now attempting to write the function expSeries which evaulates e^x using the Taylor series. This is what I have so far:
function expo = exponentialFunc(x)
terms = input('Enter the number of terms');
b = 0;
for i = 1:terms
b = x/factFunc(terms);
end
expo = b;
end
And in the main program, I have
n = exponentialFunc(4);
disp(n);
Where in this instance I am trying to find e^4. However, the output is not what expected. Does anyone have any idea where I am going wrong?
Fix to factFunc:
function fact = factFunc(n)
f = 1;
for a = 1:n
f = f*a;
end
fact = f;
end
Fix to exponentialFunc
function expo = exponentialFunc(x)
terms = input('Enter the number of terms');
b = 0;
for i = 0:terms-1
b = b + x^i/factFunc(i);
end
expo = b;
end
Example
>> exponentialFunc(4)
Enter the number of terms10
ans =
54.1541
Note exp(4) = 54.59815...
I'm facing an issue with this simple Fibonacci number generator program:
function f = fibonr(n)
f(1) = 1;
f(2) = 1;
for i = 3:n
f(i) = f(i-1) + f(i-2);
end
end
If I want to display only the n-th number of the sequence, what adjustments should I make?
function f = fibonr(n)
f = zeros(n,1); %//initialise the output array
f(1,1) = 1;
f(2,1) = 1;
for ii = 3:n
f(ii,1) = f(ii-1,1) + f(ii-2,1);
end
%//create a string with text and variables
str = sprintf('The %d th number in the Fibonacci sequence is %d',n,f(ii,1));
disp(str) %//display your output.
end
first up: don't use i as a variable. Secondly, I switched to using column vectors, since MATLAB processes them faster, as well as I initialised the array, which is way faster (hence the shiny orange wiggles below your f(i)= line).
Call your function:
output = fibonr(10);
The 10 th number in the Fibonacci sequence is 55
If you use e.g. n=20 and still want the 10th argument just call output(10)
if you want the specified output right away you can use nargin. This code will give you all of the sequence if you call fibonr(n), or you can specify a vector to get the fibonacy numbers at said positions. If you are interested in both, your specified output and all the numbers, you can call the function with:
[output, fibnumbers] = fibonr(n,v);
function [output,f] = fibonr(n,v)
f(1) = 1;
f(2) = 1;
for i = 3:n
f(i) = f(i-1) + f(i-2);
end
if nargin() > 1
output = f(v);
else
output = f;
end
This is my matlab code , I got Not enough input argument error in line 2 and i don't know how to fix it. Anyhelp ? Thanks in advance .
function [] = Integr1( F,a,b )
i = ((b - a)/500);
x = a;k = 0; n = 0;
while x <= b
F1 = F(x);
x = x + i;
F2 = F(x);
m = ((F1+F2)*i)/2;
k = k +m;
end
k
x = a; e = 0; o = 0;
while x <= (b - 2*i)
x = x + i;
e = e + F(x);
x = x + i;
o = o + F(x);
end
n = (i/3)*(F(a) + F(b) + 2*o + 4*e)
This code performs integration by the trapezoidal rule. The last line of code gave it away. Please do not just push the Play button in your MATLAB editor. Don't even think about it, and ignore that it's there. Instead, go into your Command Prompt, and you need to define the inputs that go into this function. These inputs are:
F: A function you want to integrate:
a: The starting x point
b: The ending x point
BTW, your function will not do anything once you run it. You probably want to return the integral result, and so you need to modify the first line of your code to this:
function n = Integr1( F,a,b )
The last line of code assigns n to be the area under the curve, and that's what you want to return.
Now, let's define your parameters. A simple example for F is a linear function... something like:
F = #(x) 2*x + 3;
This defines a function y = 2*x + 3. Next define the starting and ending points:
a = 1; b = 4;
I made them 1 and 4 respectively. Now you can call the code:
out = Integr1(F, a, b);
out should contain the integral of y = 2*x + 3 from x = 1 to x = 4.