将大数字类型转换为较小类型 [英] Casting a large number type to a smaller type
问题描述
我有一个很好的环顾四周,找不到一个类似的问题,如果以前被问过,请不要抱歉。
I've had a good look around and can't find a similar question so apologies if it has been asked before.
我只是玩类型和数字,我想知道是否可以保证以下行为。
如果我声明2个变量为
I'm just playing around with types and numbers and I am wondering if the following behaviour can be guaranteed. If I declare 2 variables as
unsigned char BIT_8 = 0;
unsigned short int BIT_16 = 0xFF01;
然后执行以下操作(忽略C风格的转换,除非可以影响它) / p>
and then do the following (ignoring C style cast for now, unless that can affect it?)
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";
我获得输出
BIT_16: 65281
BIT_8: 0
BIT_8 after: 1
BIT_8 after shift: 255
如果我将一个16位类型转换为一个8位类型,前导字节将丢失,这是否保证?
Is it guaranteed that if I cast a 16 bit type to an 8 bit type that the leading byte will be lost? or is it undefined and the above results are luck?
推荐答案
16位类型到8位类型,前导字节将丢失?
Is it guaranteed that if I cast a 16 bit type to an 8 bit type that the leading byte will be lost?
取决于您是使用有符号还是无符号类型(见第4.7节§2和§3):
Depends on whether you are working with signed or unsigned types (see section 4.7 §2 and §3):
如果目标类型为无符号值是与源整数一致的最小无符号整数(模2 ^ n,其中n是用于表示无符号类型的位数)。 [注意:在二进制补码表示中,这种转换是概念性的,并且位模式没有变化(如果没有截断)。]
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).]
是已签名,如果可以在目标类型(和位字段宽度)中表示,则值不变;
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。
Since you are working with unsigned types, the behavior is well-specified.
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