On Mon, 29 Jan 2007 23:05:37 -0800, ArdPy wrote:
> Hi,
>
> Pls tell me whats going on in the code snippet below:
>
>>>> n = 10
>>>> statstr = "N = ",n
>>>> type(statstr) #case1
> <type 'tuple'>
>>>> print statstr
> ('N = ', 10)
>>>> print "N = ",n #case 2
> N = 10
>
> In the first case the result is printed as a tuple and in the second
> it appears as an ordinary string.
The print statement takes one or more comma-delimited objects, and prints
each one separated with a space, and then finally prints a newline. If you
end the list with a comma, no newline is printed. So when you execute the
line:
print "N = ", n
Python prints "N = ", then a space, then 10, then a newline.
Outside of a print statement (and also an "except" statement), commas
create tuples. So when you execute the line:
statstr = "N = ", n
the right-hand side is a tuple with two items. The first item is the
string "N = " and the second item is the integer currently named n (in
your case, 10).
Do not be fooled that brackets make tuples -- they don't. With the
exception of the empty tuple () which is a special case, it is commas that
make tuples. So these two lines have identical effects:
t = 1, 2, 3
t = (1, 2, 3)
That is why these two statements print something different:
print 1,2,3
print (1,2,3)
In the second case, the brackets force the expression "1,2,3" to be
evaluated before the print statement sees it, so the print statement sees
a single argument which is a tuple, instead of three int arguments.
(But notice that the brackets still aren't creating the tuple, the commas
are. The brackets just change the order of evaluation.)
--
Steven D'Aprano
--
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