include: avoid UB with signed overflow/shift #1675
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if integer is 32-bit, and numeric literals default to int type, the following applies (from The C Standard, 6.5.7, paragraph 4 [ISO/IEC 9899:2011]): If E1 has a signed type and nonnegative value, and E1 × 2^E2 is representable in the result type, then that is the resulting value; otherwise, the behavior is undefined. which means that the only way to safely shift is unsigned, so use 1U to indicate the shifted bit is unsigned.
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aquynh
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Sep 16, 2020
if integer is 32-bit, and numeric literals default to int type, the following applies (from The C Standard, 6.5.7, paragraph 4 [ISO/IEC 9899:2011]): If E1 has a signed type and nonnegative value, and E1 × 2^E2 is representable in the result type, then that is the resulting value; otherwise, the behavior is undefined. which means that the only way to safely shift is unsigned, so use 1U to indicate the shifted bit is unsigned.
aquynh
pushed a commit
that referenced
this pull request
Sep 16, 2020
if integer is 32-bit, and numeric literals default to int type, the following applies (from The C Standard, 6.5.7, paragraph 4 [ISO/IEC 9899:2011]): If E1 has a signed type and nonnegative value, and E1 × 2^E2 is representable in the result type, then that is the resulting value; otherwise, the behavior is undefined. which means that the only way to safely shift is unsigned, so use 1U to indicate the shifted bit is unsigned.
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if integer is 32-bit, and numeric literals default to int type,
the following applies (from The C Standard, 6.5.7, paragraph 4
[ISO/IEC 9899:2011]):
If E1 has a signed type and nonnegative value, and E1 × 2^E2 is
representable in the result type, then that is the resulting value;
otherwise, the behavior is undefined.
which means that the only way to safely shift is unsigned, so
use 1U to indicate the shifted bit is unsigned.