i want to debug following simplest code in matlab and clarify why it executes always if statement
function testfile(x)
if 3<x<6
disp('in the middle of range');
else
disp('out of range');
end
end
i have used following code for debugger
echo testfile on
testfile(-2)
in the middle of range
testfile(6)
in the middle of range
why it does not execute else statement?i have used following code as a test
5<4<8
ans =
1
so does it mean that writing if statement in this style is wrong?a i understood it is same as if 5<4 || 4<8?then it makes clear for me why it executes only if statement and never reaches to else
5<4<8 is evaluated as (5<4)<8. If we resolve the expression in the parentheses first, we have 0<8, which is true. Test with 5<4==0, which evaluates to true.
What you want to do is check whether x is both bigger than 3 and smaller than 6, i.e.
3<x && x<6
Related
I have the following if else statement that created by myself in order to link to the if else statement given in second part:
m=4
if m==3
disp(true)
else
disp(false)
Second part ( this code is fix cannot be change):
if (true)
A=Hello World
else
A=Bye
If using the first part code, my output will be
A=Hello World
but my desire output is
A=Bye
Anyone one have idea to edit the first part, because now my return value in first part not able to link to my second part.
If you can't change the second part's code, I'm afraid your desire cannot be fulfilled. Or rather, I'm afraid your code won't run at all, because your perenthesis, quotes, end-statement (and arguably semicolons) are not in place.
if true
A = 'Hello World';
else
A = 'Bye';
end
This code will return A = 'Hello World', no matter what, since true is always true. If-else conditions work like this:
if (*what's in here evealuates to true*)
%do stuff
else (*if what's up there does not evaluate to true*)
%do other stuff
Clearly, true will always evaluate to true. So the above if-else condition will always return A = 'Hello World'.
You don't need two if statements in order to accomplish this task. One is more than enough to perform all what you need:
m = 4;
if (m == 3)
A = 'Hello World';
else
A = 'Bye';
end
disp(A);
A few comments concerning your code:
if statements need to be closed with an end
if (true) will always pass into the statement
the disp function doesn't assign a value, its only goal is to display it in the Command Window
in order to work with text, you have to enclose it within single quotes ' (char array) or double quotes " (string), more info here
If you posted only small excerpts of your code and you need to perform those two checks sequentially, in different parts of your script, then:
m = 4;
if (m == 3)
m_equals_3 = true;
disp('M == 3');
else
m_equals_3 = false;
disp('M ~= 3');
end
% then, somewhere else...
if (m_equals_3)
A = 'Hello World';
else
A = 'Bye';
end
% ...
I guess this is a homework exercise. You should disclose that if it’s the case.
The exercise requieres you to change the workspace such that the second bit of code evaluated the else case. This can be accomplished by changing the meaning of true. In your first bit of code, make it so that
true = flase;
Or equivalently,
true = 0;
Note that this is really bad form, if you ever do something like this outside of a homework exercise that explicitly asks you to do so, you’ll get fired or maybe even shot. You’ve been warned!
By the way, I assume that the missing quote characters around the strings and the missing ends are typos?
How do i tranform this code below to matlab? I got confused on goto statement.
do 57 i=1,10
statement 1
if(k .eq. nx) then
statement 2
go to 58
end if
57 continue
statement 3
58 continue
Using GOTO command is not considered as a good procedural programming.
Use the following program instead:
i=1;
t=true;
while (i<=10)&&t
statement1;
t=k~=nx;
i=i+1;
end
if t
statement2;
else
statement3;
end
This Fortran Snippet has a very bad code-smell to it.
But here are a few things:
true here seems to be a variable. The correct value for true in Fortran is .TRUE. (or .true.). Assuming that that variable is always .TRUE., then the code can be rewritten very easily:
statement 1
statement 2
And that's it. Your code would immediately jump out of the loop and over statement 3, so each of the statements would only be executed once.
But assuming that true is some variable or expression that has to be re-evaluated in each iteration of the loop, this is a better way:
do i = 1, 10
statement 1
if (true) exit
end do
if (true) then
statement 2
else
statement 3
end if
Now this still assumes that true is a static expression, that is that it won't change it's value between calls.
I have written a function which takes in an integer (int8) as one of the inputs (called iscool). The function runs a while loop and I insert an if-check inside it to break out of the loop. The if-check checks the value of the iscool variable as well and sets the value of imarch to 0 to get out of loop. So basically, the code is something like this.
% Code_snippet
while (imarch == 1)
<some procedures not modifying iscool>
if ((iscool == 0) && (<other condition 1>) && (<other condition 2>))
imarch = 0;
elseif ((iscool == 1) && (<other condition 3>) && (<other condition 4>))
imarch = 0;
end
disp (strcat('Cooling index is: ',num2str(iscool)));
end
The output of the disp command in the first while-loop execution is 0 (which is the input), but it changes to 1 in the subsequent iteration and stays so after that. I have tried removing the if-elseif-end check and the value of iscool stays intact in that case, but I need to have the check in order to be able to get out of the loop. Any sort of help, particularly an insight into why the value might be changing would be great help. Thanks.
I have been trying to print some complex numbers. If the complex number (modTrace) is like 'a-ib' then my code:
modTrace
v = [real(modTrace(:)) imag(modTrace(:))].';
fprintf(fileID,'%e+%ei\n',v);
gives the output as:
2.355387e-13+-7.217925e-13i
To avoid the extra + sign in front of the negative imaginary piece I write:
v = [real(modTrace(:)) imag(modTrace(:))].';
if imag(modTrace(:))>0
fprintf(fileID,'%e+%ei\n',v);
else
fprintf(fileID,'%e%ei\n',v);
end
Now in the output the 'if' is not being executed. So if I have a complex number 'a+ib' it prints
'a bi' and if a complex number is 'a-ib' it prints 'a-bi' according to the else statement.
if I then modify the code like:
v = [real(modTrace(:)) imag(modTrace(:))].';
if imag(modTrace(:))<0
fprintf(fileID,'%e%ei\n',v);
else
fprintf(fileID,'%e+%ei\n',v);
end
then again the 'if' statement in not being executed. So for a complex number 'a-ib' the output is 'a+-bi' and the 'else' statement in being executed correctly.
Could any body please help me to find the output in the correct form?
Thanks.
I think that in this case you are better off using a loop:
for k = 1:length(modTrace)
if imag(modTrace(k))>0
fprintf(fileID,'%e+%ei\n',real(modTrace(k)), imag(modTrace(k)));
else
fprintf(fileID,'%e%ei\n',real(modTrace(k)), imag(modTrace(k)));
end
end
You can't vectorize it that easily because if you pass a vector to if it only evaluates true when (http://www.mathworks.com/help/matlab/ref/if.html):
the result is nonempty and contains all nonzero elements (logical or
real numeric). Otherwise, the expression is false.
Here is my code:
function [im,sindx,end1]=alln(im,i,j,secret,sindx,end1)
slen=length(secret);
p=im(i,j);
neigh= [im(i-1,j) im(i+1,j) im(i,j-1) im(i,j+1) im(i-1,j-1) im(i+1,j-1) im(i-1,j+1) im(i+1,j+1)];
minpix = min (neigh)
maxpix = max (neigh)
if minpix < p < maxpix
lowlim = minpix+1;
highlim = maxpix-1;
range = highlim-lowlim+1;
nbits=floor(log2(abs(range)));
if sindx+nbits-1>slen
end1=1;
return
end
for k=1:nbits
bin(k)=secret(sindx+k-1);
end
b = bin2dec(bin);
newvalue1 = abs (minpix + b);
newvalue2 = abs (maxpix - b);
if abs(p-newvalue1)<= abs(p-newvalue2)
im(i,j) = newvalue1;
else
im(i,j) = newvalue2;
end
sindx=sindx+nbits;
end
end
My main program calls this function. When I run the program, I get the following error message:
??? Undefined function or variable "bin".
Error in ==> alln at 34
b = bin2dec(bin);
I know there are many experts for whom this is not a problem at all. I am new to MATLAB. Please guys, show me the way, which type of modification in the code can overcome this problem?
First of all, are there some lines missing from the file? Perhaps you've stripped some comments from the top? Because the error message says that
b = bin2dec(bin);
is line 34, but it's line 22 in the code you present.
OK, that aside...
The error message says that 'bin' isn't defined, but I see that it's being set on the line...
bin(k)=secret(sindx+k-1);
That suggests to me that THAT line isn't being run.
I see that that bin = ... line is inside of a 'for' loop, so I suspect that the for loop is run zero times, meaning that 'bin' never gets defined. What is nbits? Is it 1, or perhaps less than 1? THAT would prevent the loop from running at all.
Try removing the semicolon from the end of the
nbits=floor(log2(abs(range)));
line and run your code again.
Leaving off the semicolon will force the value of nbits to be printed in the Command Window. I bet you'll find that it's 1 or less. If that's the case, then start looking at HOW nbits is calculated, and I bet you'll find the problem.
At what input arguments to the function alln, are you getting the error?
Lets suppose that nbits is 0, then the following loop will not run:
for k=1:nbits
bin(k)=secret(sindx+k-1);
end
So, bin will be undefined. So, the error happens. This is one of the cases where the error can happen. There are many such possible cases.