datatypes

조회 수: 12 (최근 30일)
sheen
sheen 2011년 5월 13일
double e1,e2,e;
e1 = 107; e2 = e1 * 339;
disp(e2/e1)
error: The string being specified was neither 'single' nor 'double' ??? Undefined function or variable "e2".
Error in ==> ef at 1 float e1,e2,e;
ans =
101 49
??? Undefined function or variable "e2".
Error in ==> ef at 1 double e1,e2,e;
it is also giving problem with int and float?? what should i do to deal with this error?

채택된 답변

Walter Roberson
Walter Roberson 2011년 5월 13일
Your command
double e1,e2,e;
is equivalent to
double('e1')
e2
e
'e1' is a string, which is an array of character, and applying double to the array of character returns the numeric values of each of the characters: that happens to be 101 for 'e' and 49 for '2'. For more information on this, please see Command vs Function syntax
In MATLAB, one does not declare variables as being of a particular type: one just assigns values and the variable assumes the type of the values if the variable appears by itself (without any kind of indexing) in an assignment syntax.
  댓글 수: 26
sheen
sheen 2011년 5월 21일
d (P1, P2) =
M j j-1
∑ Q( (∑ (u )/T) - Q ((∑ (u )/T) * d v (P1,P2)
(j=1) (k=1) k (k=1) k j
i have written code for this eq.
j=1 is subscrtipt of first summation.and M is superscript of first summation and so on.i hope u will understand my problem now.for u with k as subscrit , i have taken size, modes and cost drivers.please tell me how to correct my code?
sheen
sheen 2011년 5월 21일
j is superscript of second summation , j-1 is superscript of third summation.k=1 is subscript of second summation,k is of u, k=1 is of third summation and j issubscript of v.

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추가 답변 (1개)

Sean de Wolski
Sean de Wolski 2011년 5월 13일
You don't need to declare them as double; it's automatic.
e1 = 107;
e2 = e1 * 339;
disp(e2/e1)
  댓글 수: 6
sheen
sheen 2011년 5월 18일
how to call fuzzy routine in this code ?
sheen
sheen 2011년 5월 21일
how to use FIS output in this above similarity measure code?

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