x = x + ";" adds 2 semicolons to x?

[Geminias]

Python:

for i in range(len(self.functions[0].b)):             self.functions[0].b[i] = (self.functions[0].b[i] + ";")             DebugOut(self.functions[0].b[i])

Seriously... why does self.functions[0].b[i] end with 2 semicolons although I am only adding one. I am also sure there is no semicolon originally. But for some reason it actually adds 2 although I am clearly saying to only add 1.

This has me baffled.

[Zeroth]

There's absolutely no reason why it would do that. What is in the variable named 'b'? If they are strings, then there should be no reason for that. If they're objects masquerading as strings, maybe there's a bug in some of the overridden functions?

Of course, maybe your debugout function is adding the semi-colons itself?

Without the rest of the code, we really can't diagnose the problem.

code:

>>> a=range(10) >>> for i in range(len(a)): ...     a[i]=(str(a[i])+';') ... >>> a ['0;', '1;', '2;', '3;', '4;', '5;', '6;', '7;', '8;', '9;']

[[Gandalf]]

Indeed, just for the sake of double checking... Assuming self.functions[0].b is a list of strings, the following is an exact replica of your code without the DebugOut() calls.

Python:

a = ["hello", "hi", "bye"] for i in range(len(a)):     a[i] = (a[i] + ";")

Quote:

['hello;', 'hi;', 'bye;']

[octopi]

I would try adding a "" to the end, or another value such as "X", and seeing if you receive the proper output. If it is part of an object it could be that it does some extra stuff which is adding an extra semicolon when you change the value.

[wtd]

Zeroth @ Wed Feb 18, 2009 12:44 pm wrote:

There's absolutely no reason why it would do that. What is in the variable named 'b'? If they are strings, then there should be no reason for that. If they're objects masquerading as strings, maybe there's a bug in some of the overridden functions?

Of course, maybe your debugout function is adding the semi-colons itself?

Without the rest of the code, we really can't diagnose the problem.

code:

>>> a=range(10) >>> for i in range(len(a)): ...     a[i]=(str(a[i])+';') ... >>> a ['0;', '1;', '2;', '3;', '4;', '5;', '6;', '7;', '8;', '9;']

I think you meant:

code:

>>> map(lambda x: str(x) + ";", range(10)) ['0;', '1;', '2;', '3;', '4;', '5;', '6;', '7;', '8;', '9;'] >>>

[Zeroth]

wtd @ Tue Feb 17, 2009 10:30 pm wrote:

Zeroth @ Wed Feb 18, 2009 12:44 pm wrote:

There's absolutely no reason why it would do that. What is in the variable named 'b'? If they are strings, then there should be no reason for that. If they're objects masquerading as strings, maybe there's a bug in some of the overridden functions?

Of course, maybe your debugout function is adding the semi-colons itself?

Without the rest of the code, we really can't diagnose the problem.

code:

>>> a=range(10) >>> for i in range(len(a)): ...     a[i]=(str(a[i])+';') ... >>> a ['0;', '1;', '2;', '3;', '4;', '5;', '6;', '7;', '8;', '9;']

I think you meant:

code:

>>> map(lambda x: str(x) + ";", range(10)) ['0;', '1;', '2;', '3;', '4;', '5;', '6;', '7;', '8;', '9;'] >>>

I think I meant something readable I only use maps and lambdas either A) when the normal flow controls can't do what I want well enough or B) for efficiency, which I only do after I get the code working in the first place. And I still leave the old structure there in comments so people can see what it does easily. But to each their own!

[wtd]

One big reason I prefer Ruby to Python: Python is overly statement=oriented, and treats expressions as a necessary evil.