Return a zero-filled register with uint8_t - #10
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Apparently, gcc expects a uint8_t to be zero-padded if it's
loaded into a register. A code that fails looks like this:
uint8_t addr;
uint64_t full_906 = (uint64_t)BUILD_GEN << (32 + 24);
full_906 |= ((uint64_t)pfI2cCfgs[2].slaveAddress) << (32 + 16);
addr = HW_get_8bit_reg(pfI2cCfgs[2].baseAddress, ADDRESS);
full_906 |= ((uint64_t)addr) << (32 + 8);
At this point, the value of full_906 is 0xFFFFFF8400000000
and the field of BUILD_GEN is destroyed. Observe,
there's no signed int sneaking in through promotion.
The sign-extension happens inside the assembly code.
Therefore, we fix it.
Signed-off-by: Pete Zaitcev <zaitcev@yahoo.com>
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I found that the following code that fails:
At this point, the value of full_906 is 0xFFFFFF8400000000 and the field of BUILD_GEN is destroyed. Observe,
there's no signed int sneaking in through promotion. The sign-extension happens inside the assembly code.
Apparently, gcc expects a uint8_t to be zero-padded if it's loaded into a register. It does not add any additional masking before shifting.
An easy fix is to use the
lbuinstruction instead of thelb.I don't know if this applies to shorts and ints, so for safety's sake, I didn't try to change those accessors.