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13b13d1fdd
This includes making int("01") parse in base 10 like standard Python.
When a base of 0 is specified it means auto-detect based on the prefix, and
literals begining with 0 (except when the literal is all 0's) like "01" are
then invalid and now throw an exception.
The new error message is different from CPython. It says e.g.,
`SyntaxError: invalid syntax for integer with base 0: '09'`
Additional test cases were added to cover the changed & added code.
Co-authored-by: Damien George <damien@micropython.org>
Signed-off-by: Jeff Epler <jepler@gmail.com>
94 lines
1.5 KiB
Python
94 lines
1.5 KiB
Python
# test the lexer
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try:
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eval
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exec
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except NameError:
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print("SKIP")
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raise SystemExit
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# __debug__ is a special symbol
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print(type(__debug__))
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# short input
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exec("")
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exec("\n")
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exec("\n\n")
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exec("\r")
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exec("\r\r")
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exec("\t")
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exec("\r\n")
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exec("\nprint(1)")
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exec("\rprint(2)")
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exec("\r\nprint(3)")
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exec("\n5")
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exec("\r6")
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exec("\r\n7")
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print(eval("1"))
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print(eval("12"))
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print(eval("123"))
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print(eval("1\n"))
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print(eval("12\n"))
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print(eval("123\n"))
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print(eval("1\r"))
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print(eval("12\r"))
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print(eval("123\r"))
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# line continuation
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print(eval("'123' \\\r '456'"))
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print(eval("'123' \\\n '456'"))
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print(eval("'123' \\\r\n '456'"))
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print(eval("'123'\\\r'456'"))
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print(eval("'123'\\\n'456'"))
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print(eval("'123'\\\r\n'456'"))
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# backslash used to escape a line-break in a string
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print('a\
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b')
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# lots of indentation
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def a(x):
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if x:
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if x:
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if x:
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if x:
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if x:
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if x:
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if x:
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if x:
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if x:
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if x:
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if x:
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if x:
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if x:
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if x:
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if x:
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print(x)
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a(1)
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# badly formed hex escape sequences
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try:
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exec(r"'\x0'")
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except SyntaxError:
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print("SyntaxError")
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try:
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exec(r"b'\x0'")
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except SyntaxError:
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print("SyntaxError")
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try:
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exec(r"'\u000'")
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except SyntaxError:
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print("SyntaxError")
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try:
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exec(r"'\U0000000'")
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except SyntaxError:
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print("SyntaxError")
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# Properly formed integer literals
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print(eval("00"))
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# badly formed integer literals
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try:
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eval("01")
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except SyntaxError:
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print("SyntaxError")
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