The getenv() function is inherently not reentrant because it returns a value pointing to static data.
In fact, for higher performance of getenv(), the implementation could also maintain a separate copy of the environment in a data structure that could be searched much more quickly (such as an indexed hash table, or a binary tree), and update both it and the linear list at environ when setenv() or unsetenv() is invoked.
So the address returned by getenv is not necessarily from the environment.
Process memory layout;
(source: duartes.org)
(source: cloudfront.net)
Memory map
import os
def mem_map():
path_hex = hex(id(os.getenv('PATH'))).rstrip('L')
path_address = int(path_hex, 16)
for line in open('/proc/self/maps'):
if 'stack' in line:
line = line.split()
first, second = line[0].split('-')
first, second = int(first, 16), int(second, 16)
#stack grows towards lower memory address
start, end = max(first, second), min(first, second)
print('stack:\n\tstart:\t0x{}\n\tend:\t0x{}\n\tsize:\t{}'.format(start, end, start - end))
if path_address in range(end, start+1):
print('\tgetenv("PATH") ({}) is in the stack'.format(path_hex))
else:
print('\tgetenv("PATH") ({}) is not in the stack'.format(path_hex))
if path_address > start:
print('\tgetenv("PATH") ({}) is above the stack'.format(path_hex))
else:
print('\tgetenv("PATH") ({}) is not above the stack'.format(path_hex))
print('')
continue
if 'heap' in line:
line = line.split()
first, second = line[0].split('-')
first, second = int(first, 16), int(second, 16)
#heap grows towards higher memory address
start, end = min(first, second), max(first, second)
print('heap:\n\tstart:\t0x{}\n\tend:\t0x{}\n\tsize:\t{}'.format(start, end, end - start))
if path_address in range(start, end+1):
print('\tgetenv("PATH") ({}) in the heap'.format(path_hex))
else:
print('\tgetenv("PATH") ({}) is not in the heap'.format(path_hex))
print('')
Output;
heap:
start: 0x170364928
end: 0x170930176
size: 565248
getenv("PATH") (0xb74d2330) is not in the heap
stack:
start: 0x0xbffa8000L
end: 0x0xbff86000L
size: 139264
getenv("PATH") (0xb74d2330) is not in the stack
getenv("PATH") (0xb74d2330) is not above the stack
Environment is above the stack. So its address should be higher than the stack. But the address id shows is not in the stack, not in the heap and not above the stack. Is it really an address? or my calculation is wrong!
Here's the code to check where an object lies in memory.
def where_in_mem(obj):
maps = {}
for line in open('/proc/self/maps'):
line = line.split()
start, end = line[0].split('-')
key = line[-1] if line[-1] != '0' else 'anonymous'
maps.setdefault(key, []).append((int(start, 16), int(end, 16)))
for key, pair in maps.items():
for start, end in pair:
# stack starts at higher memory address and grows towards lower memory address
if 'stack' in key:
if start >= id(obj) >= end:
print('Object "{}" ({}) in the range {} - {}, mapped to {}'.format(obj, hex(id(obj)), hex(start), hex(end), key))
continue
if start <= id(obj) <= end:
print('Object "{}" ({}) in the range {} - {}, mapped to {}'.format(obj, hex(id(obj)), hex(start), hex(end), key))
where_in_mem(1)
where_in_mem(os.getenv('PATH'))
Output;
Object "1" (0xa17f8b0) in the range 0xa173000 - 0xa1fd000, mapped to [heap]
Object "/usr/local/sbin:/usr/local/bin:/usr/sbin:/usr/bin:/sbin:/bin:/usr/games:/usr/local/games" (0xb74a1330L) in the range 0xb7414000L - 0xb74d6000L, mapped to anonymous
What's anonymous in the above output?
It is also possible to create an anonymous memory mapping that does not correspond to any files, being used instead for program data. In Linux, if you request a large block of memory via malloc(), the C library will create such an anonymous mapping instead of using heap memory. ‘Large’ means larger than MMAP_THRESHOLD bytes, 128 kB by default and adjustable via mallopt().
Anatomy of a Program in Memory
So the os.environ['PATH'] is in the malloced region.
