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• Consider the following three statements. Do they change the value printed for A?

  1. A = "spam"
     
  2. B = A
     
  3. B = "shrubbery"

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• Consider these three statements. Do they change the printed value of A?

  1. A = ["spam"]
     
  2. B = A
     
  3. B[0] = "shrubbery"

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• How about these—is A changed now?

  1. A = ["spam"]
     
  2. B = A[:]
     
  3. B[0] = "shrubbery"

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关键词：knowledge python Edge test Your statements following printed change about

• No: A still prints as "spam". When B is assigned to the string "shrubbery", all that happens is that the variable B is reset to point to the new string object. A and Binitially share (i.e., reference/point to) the same single string object "spam", but two names are never linked together in Python. Thus, setting B to a different object has no effect on A. The same would be true if the last statement here were B = B + 'shrubbery', by the way—the concatenation would make a new object for its result, which would then be assigned to B only. We can never overwrite a string (or number, or tuple) in place, because strings are immutable.

• Yes: A now prints as ["shrubbery"]. Technically, we haven’t really changed either A or B; instead, we’ve changed part of the object they both reference (point to) by overwriting that object in place through the variable B. Because A references the same object as B, the update is reflected in A as well.

• No: A still prints as ["spam"]. The in-place assignment through B has no effect this time because the slice expression made a copy of the list object before it was assigned to B. After the second assignment statement, there are two different list objects that have the same value (in Python, we say they are ==, but not is). The third statement changes the value of the list object pointed to by B, but not that pointed to by A.