Emulate a printf() C-statement in Python???

Eric Brunel eric.brunel at nospam-pragmadev.com
Thu Mar 19 14:08:35 CET 2009


Mr. Z wrote:
> I'm trying emulate a printf() c statement that does, for example
>
> char* name="Chris";
> int age=30;
> printf("My name is %s", name);
> printf("My name is %s and I am %d years old.", %s, %d);
>
> In other words, printf() has a variable arguement list the we
> all know.
>
> I'm trying to do this in Python...
>
> class MyPrintf(object):
>     # blah, blah
>      def myprintf(object, *arg):
>           # Here I'll have to know I NEED 2 arguments in format string
> arg[0]
>           print arg[0] % (arg[1], arg[2])

No you don't: in Python, the % operator on strings expects a tuple as second
operand, and args is already a tuple. So:
print args[0] % arhgs[1:]
will do what you want.

Some stylistic issues:
- You don't have to put everything in a class in Python. Your myprintf looks a
lot like a function to me, so you'll probably want to define it as such.
- Don't use object as the first parameter to methods! object is the top-level
class for the whole class hierarchy, so you certainly don't want to obscure it
with a variable. For methods, use the traditional self as first parameter.
- In your function myprintf, the format string seems to be a required
argument. If it is, you might want to define as such too:

def myprintf(format_string, *args):
  print format_string % args

So now, as you can see, redefining printf in Python is not really
interesting...

HTH anyway.
 - Eric -



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