| 1） int32_t xxx1 = 33 & 31; int32_t xxx2 = 64 & 31; int32_t xxx3 = 16 & 31; int32_t xxx4 = 100 & 31; std::cout << xxx1<<","<<xxx2<<","<<xxx3<<","<<xxx4 <<std::endl; 2）结果 1,0,16,4 3）分析 31表示成2进制为11111, 33表示成2进制为100001 64表示成2进制为1000000 4） int32_t xxx1 = 63 & 31; int32_t xxx2 = 64 & 31; int32_t xxx3 = 16 & 31; int32_t xxx4 = 100 & 31; 结果
31,0,16,4
5） int32_t xxx1 = 63 & 31; int32_t xxx2 = 64 & 31; int32_t xxx3 = 94 & 31; int32_t xxx4 = 97 & 31; std::cout << xxx1<<","<<xxx2<<","<<xxx3<<","<<xxx4 <<std::endl; 31,0,30,1
5）相当于对32取余数。 |
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