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does parsing expression in parser.parse_expr() act the same as parsing outside? I called parse_expr (with no arguments) with current token of the first num1 and it kept on parsing until the right parenthesis of asm!
Does the code make use of any (1.49) nightly feature ?
Nightly
Godbolt link
No response
Actual behavior
Currently the inline assembly uses 2 identifier token checks, one for num1 before the =>, one for num1 after the =>
Expected behavior
I expected to be able to use parse_expr for both identifier, meaning I would want to call parse_expr(), then check for => , then call parse_expr() again without parse_expr eating up all token from the token stream
Summary
From Jun 5th,
badumbatish:
Reproducer
I tried this code:
Does the code make use of any (1.49) nightly feature ?
Godbolt link
No response
Actual behavior
Currently the inline assembly uses 2 identifier token checks, one for num1 before the
=>
, one for num1 after the=>
Expected behavior
I expected to be able to use parse_expr for both identifier, meaning I would want to call parse_expr(), then check for
=>
, then call parse_expr() again without parse_expr eating up all token from the token streamGCC Version
#3098
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