Conversion error when accessing bits to set a bit field

Question

We are using bit fields to represent elements of a register read from a device.

#include <cstdint> struct Register { uint8_t field : 1; }; int main() { uint8_t byte{0}; // Read from a device Register r; r.field = (byte >> 5) & 0x1; // access bit 5 r.field = (byte >> 7) & 0x1; // access bit 7, warns }

We are also using the flag -Werror=conversion. For some reason, accessing bit 0 through 6 compiles without warning. However, accessing bit 7 warns for the conversion error: conversion from 'unsigned char' to 'unsigned char:1' may change value [-Werror=conversion].

Any ideas why this might be? Or how to right it in a way that will not warn of a conversion error?

Example here, https://godbolt.org/z/Ghd5ndnKd

You told the compiler that you'd like narrowing conversion warnings and that's what it gave you. If you don't like it, get rid of -Wconversion, it was always unreliable and prone to false positives. — Lundin
– Lundin, Commented Jul 11, 2024 at 13:42
What if byte is a wider type when you select a bit from it? — Davis Herring
– Davis Herring, Commented Jul 11, 2024 at 13:47
Potential duplicate of Son of GCC conversion warning when assigning to a bitfield. Most likely a compiler bug, int shift = 7; r.field = (byte >> shift) & 0x1; compiles — ipodtouch0218
– ipodtouch0218, Commented Jul 11, 2024 at 13:48
bitfields probably don't do what you think they do. They do not model addressable bits, but only create a variable that behaves as if it had N bits. Further the way bitfields behave is compiler specific. My advice, don't use them and do all your bit arithmetic manually ( — Pepijn Kramer
– Pepijn Kramer, Commented Jul 11, 2024 at 15:17
@PepijnKramer OP is actually using the bit-field without relying on any underlying representation. Their usage is completely portable. — user17732522
– user17732522, Commented Jul 11, 2024 at 17:31

Drew Dormann · Accepted Answer · 2024-07-12 15:28:48Z

9

It's indeed odd that you are getting 1 warning instead of 2 or 0.

Particularly suspicious wording conversion from 'unsigned char', because your (byte >> 7) & 0x1 has type int.

But since you're asking how to remove the warning, cast the value to a bool.

r.field = ((byte >> 7) & 0x1) != 0; // access bit 7

or

r.field = bool((byte >> 7) & 0x1); // access bit 7

edited Jul 12, 2024 at 15:28

answered Jul 11, 2024 at 13:41

Drew Dormann

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3 Comments

Pete Becker Over a year ago

Thus demonstrating that dodging warnings on code whose meaning is well defined is often counter-productive. <g> Yes, these work, but both are misleading and inelegant. Nevertheless, +1.

463035818_is_not_an_ai Over a year ago

perhaps add a comment along the line of // convert to bool to silence -Werror=conversion

Neil Over a year ago

Somehow this actually generates better code at -O0.

Ted Lyngmo · Accepted Answer · 2024-08-01 14:42:52Z

It's a diagnostic bug. You can of course work around it, but rather than recognizing the bad diagnostics message, I suggest that you suppress it:

// Suppress error: // "conversion from 'unsigned char' to 'unsigned char:1' may change value" #pragma GCC diagnostic ignored "-Wconversion" r.field = (byte >> 7) & 0x1; #pragma GCC diagnostic pop

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