Help me understand this

[Steven D'Aprano]

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