https://gcc.gnu.org/bugzilla/show_bug.cgi?id=97833
Bug ID: 97833
Summary: -Wconversion behaves erratic
Product: gcc
Version: 10.2.0
Status: UNCONFIRMED
Severity: normal
Priority: P3
Component: c
Assignee: unassigned at gcc dot gnu.org
Reporter: sven.koehler at gmail dot com
Target Milestone: ---
Created attachment 49559
--> https://gcc.gnu.org/bugzilla/attachment.cgi?id=49559&action=edit
non-working example
Find attached an example for which -Wconversion behaves uncomprehensible.
Why does it yields warning for test2 but not for test1 and test3?
This happens with gcc for 32bit arm and gcc for x86_64.
In all 3 functions, we have 2 shift operations. The operand is uint16_t, which
is promoted to int. The result of the shift operations is cast to uint16_t. So
the operands of the bit-wise or are again uint16_t. So both operands of the
bitwise or are promoted to int.
So basically, in all 3 cases the code is returning an int. However, -W
conversion warns only in 1 case.
Also, why does it matter whether x is shifted by 0 or 1 ? Why does a shift by 0
result in an error, and a shift by 1 does not?
Why does it matter whether x and y are originally uint8_t being cast to
uint16_t (test2) or a uint16_t (test3) originally? In both cases, the result of
the shifts is cast to uint16_t.
Is gcc trying to keep track of the range of the individual expressions? Is gcc
somehow failing when the a shift by 0 occurs? I believe that a shift by zero is
defined behavior.
