位字段并集的大小，其成员数多于其大小(Size of bit-field union which has more members than its size)
#include <stdio.h> union mix { unsigned char a1:1; unsigned char a2:4; unsigned char a3:4; unsigned char a4:1; unsigned char a5:4; unsigned char a6:4; unsigned char a7:1; unsigned char a8:4; unsigned char a9:4; unsigned char a10:1; unsigned char a11:4; unsigned char a12:4; }; int main() { printf("Sizeof mix = %d bytes\n", sizeof(union mix)); return 0; }输出为1个字节 。
超过8位的位域成员会发生什么？ 显然，我仍然可以为任何位字段成员设置并获取正确的值。
UPDATE
谢谢你清除我的困惑。 一个问题：这些位以什么顺序存储在内存中？ 假设它是小端存储器，那么0xabcd将作为0xd，0xc，0xb，0xa存储在存储器中。
- m.a1会成为0xd OR oxa的一部分吗？
- 它是0xd（或0xa）的第3位还是第0位？
#include <stdio.h> union mix { unsigned char a1:1; unsigned char a2:4; unsigned char a3:4; unsigned char a4:1; unsigned char a5:4; unsigned char a6:4; unsigned char a7:1; unsigned char a8:4; unsigned char a9:4; unsigned char a10:1; unsigned char a11:4; unsigned char a12:4; }; int main() { printf("Sizeof mix = %d bytes\n", sizeof(union mix)); return 0; }The output is 1 byte.
What happens to bit-field members which exceed 8 bits ? Apparently I can still set and get correct values for any bit-field member.
UPDATE
Thank you for clearing my confusion. One side question: in what order do the bits get stored in memory ? Assuming it is little endian memory so 0xabcd will be stored in memory as 0xd, 0xc, 0xb, 0xa.
- Will m.a1 be part of 0xd OR oxa?
- Will it be bit 3 or bit 0 of 0xd(or 0xa)?
原文：https://stackoverflow.com/questions/40006207
更新时间：2024-05-03 14:05
最满意答案
这应该工作（
meds是带有dicts的列表）：tmp = {} for med in meds: name = med.pop('medication_name') tmp.setdefault(name, []).append(med) res = [{'medication_name': k, 'trend': v} for k, v in tmp.items()]结果：
>>> res [{'medication_name': 'Avonex', 'trend': [{'timeline': '2015-Jun', 'total_prescriptions': 1}, {'timeline': '2015-Jul', 'total_prescriptions': 0}]}, {'medication_name': 'Actemra IV', 'trend': [{'timeline': '2015-Jun', 'total_prescriptions': 1}, {'timeline': '2015-Jul', 'total_prescriptions': 0}]}, {'medication_name': 'Adempas', 'trend': [{'timeline': '2015-Jun', 'total_prescriptions': 0}, {'timeline': '2015-Jul', 'total_prescriptions': 0}, {'timeline': '2015-Aug', 'total_prescriptions': 0}]}]注意
这会修改原始数据。 您可以使用
copy.deepcopy()来避免这种情况：import copy tmp = {} for med in copy.deepcopy(meds): ...This should work (
medsis the list with dicts):tmp = {} for med in meds: name = med.pop('medication_name') tmp.setdefault(name, []).append(med) res = [{'medication_name': k, 'trend': v} for k, v in tmp.items()]Result:
>>> res [{'medication_name': 'Avonex', 'trend': [{'timeline': '2015-Jun', 'total_prescriptions': 1}, {'timeline': '2015-Jul', 'total_prescriptions': 0}]}, {'medication_name': 'Actemra IV', 'trend': [{'timeline': '2015-Jun', 'total_prescriptions': 1}, {'timeline': '2015-Jul', 'total_prescriptions': 0}]}, {'medication_name': 'Adempas', 'trend': [{'timeline': '2015-Jun', 'total_prescriptions': 0}, {'timeline': '2015-Jul', 'total_prescriptions': 0}, {'timeline': '2015-Aug', 'total_prescriptions': 0}]}]Note
This modifies the original data. You can use
copy.deepcopy()to avoid this:import copy tmp = {} for med in copy.deepcopy(meds): ...
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