# Operands to the || and && operators must be convertible to logical scalar values.

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Thayumanavan on 10 Apr 2014
Answered: sai m on 3 Mar 2015
%Code x=[1.01 ;2.23; 3.456 ;4.11; 5.897 ;6.234; 7.456; 8.567; 9.110; 10.333];
a=0;
if((x>=5.2)&&(x<=8.50))
disp('Inside if loop');
a=a+1;
end
a
I want the no of values greater than some value and less than some value in a 1D array. If i am going for if statement with sysntax if((x(:,1)>=5.2)&&(t(:,1)<=8.50)) i am getting Operands to the and && operators must be convertible to logical scalar values error.
If i use if((x(:,1)>=5.2)&(t(:,1)<=8.50)) flow is not going inside the loop and i am not able to get the count as 3 .
Image Analyst on 10 Apr 2014
You don't have a loop at all! There is no "for" and an "if" block is not a loop.

Image Analyst on 10 Apr 2014
Try this:
x=[1.01 ;2.23; 3.456 ;4.11; 5.897 ;6.234; 7.456; 8.567; 9.110; 10.333];
for k = 1 : length(x)
if x(k) >= 5.2 && x(k) <=8.50
fprintf('x(%d) = %f and is in range so we are inside if block.\n', k, x(k));
end
end
In command window:
x(5) = 5.897000 and is in range so we are inside if block.
x(6) = 6.234000 and is in range so we are inside if block.
x(7) = 7.456000 and is in range so we are inside if block.

Marta Salas on 10 Apr 2014
Edited: Marta Salas on 10 Apr 2014
find returns the indexes of the values that holds the condition 5.2=< x >=8.50
index = find(x>=5.2 & x<=8.50)
To know the number of values, you can check the size of index
length(index)
##### 2 CommentsShowHide 1 older comment
Marta Salas on 10 Apr 2014
That code works for me.
I would suggest you to use fprintf instead of disp to print comments with variables
fprintf('the count is %d \n',a);

Horia on 18 May 2014
Try: if condition 1 do something else if condition 2 do the samething; else do something else break; end; end;

sai m on 3 Mar 2015
error ??? Operands to the and && operators must be convertible to logical scalar values.