c++ - 将大数字类型转换为较小类型

Question

我环顾四周，找不到类似的问题，如果以前有人问过，请道歉。

我只是在玩类型和数字，我想知道是否可以保证以下行为。如果我将 2 个变量声明为

unsigned char BIT_8 = 0;
unsigned short int BIT_16 = 0xFF01;

然后执行以下操作(暂时忽略 C 样式转换，除非这会影响它？)

cout << "BIT_16: " << BIT_16 << "\n";
cout << "BIT_8: " << (int)BIT_8 << "\n";
BIT_8 = BIT_16;
cout << "BIT_8 after: " << (int)BIT_8 << "\n";
BIT_8 = BIT_16 >> 8;
cout << "BIT_8 after shift: " << (int)BIT_8 << "\n";

我得到了输出

BIT_16: 65281
BIT_8: 0
BIT_8 after: 1
BIT_8 after shift: 255

如果我将 16 位类型转换为 8 位类型，是否保证会丢失前导字节？还是未定义，上面的结果是运气？

Answer 1

Is it guaranteed that if I cast a 16 bit type to an 8 bit type that the leading byte will be lost?

取决于您使用的是有符号类型还是无符号类型(请参阅第 4.7 节第 2 节和第 3 节):

If the destination type is unsigned, the resulting value is the least unsigned integer congruent to the source integer (modulo 2^n where n is the number of bits used to represent the unsigned type). [Note: In a two's complement representation, this conversion is conceptual and there is no change in the bit pattern (if there is no truncation).]

If the destination type is signed, the value is unchanged if it can be represented in the destination type (and bit-field width); otherwise, the value is implementation-defined.

由于您使用的是无符号类型，因此行为是明确指定的。