New submission from Emmanuel Arias <[email protected]>:
Hi,
I can see that on parsetok.c there is the next block:
```
#ifdef PY_PARSER_REQUIRES_FUTURE_KEYWORD
#if 0
static const char with_msg[] =
"%s:%d: Warning: 'with' will become a reserved keyword in Python 2.6\n";
static const char as_msg[] =
"%s:%d: Warning: 'as' will become a reserved keyword in Python 2.6\n";
static void
warn(const char *msg, const char *filename, int lineno)
{
if (filename == NULL)
filename = "<string>";
PySys_WriteStderr(msg, filename, lineno);
}
#endif
#endif
```
I think that the #if 0 block can be remove safely, right?
I don't know if the #ifdef PY_PARSER_REQUIRES_FUTURE_KEYWORD we can remove.
----------
messages: 357387
nosy: eamanu
priority: normal
severity: normal
status: open
title: #if 0 block on parsetok.c
versions: Python 3.9
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Python tracker <[email protected]>
<https://bugs.python.org/issue38903>
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