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memset() treats the passed buffer as a char* array, assigning to every 1-byte of the array the value. So for a single 4-byte int32_t element, it is assigning bytes 0, 1, 2 and 3 of it to -1. It happens that -1 is 0xFF, so in the end the uint32_t is set to 0xFFFFFFFF, which is -1 in the two's complement, so the memset() actually produces the correct result in the end, assuming the platform uses two's complement integers. Assigning it in the loop is less error-prone, as using memset() on non-1-byte wide arrays with a non-zero value is fishy, and it is more portable as we don't have to assume the use of two's complement. It looks like in a future version of the C standard, C23, two's complement is the only integer format in C23 (thanks to @robinlinden on IRC for pointing that out), so perhaps we shouldn't be as concerned with the portability here? Though @iphydf says that it's still a good idea to use a for-loop for this case. |
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