C -Cuzz by C类型促销从短到INT
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I have a question that needs guidance from any expert:
As a value with
shorttype is passed as an argument toprintf()function, it'll be automatically promoted tointtype, that is why theprintf()function will see the value asinttype instead ofshorttype.So basically
shorttype is 16-bits wide, which is0000000000000000whileinttype is 32-bits wide, which is00000000000000000000000000000000.Let's say I declare a variable call
numwithshorttype and initialise it with a value of -32, that means the most significant bits of theshorttype will be1, which is0000000011100000.When I pass this value to
printf(), it'll be converted tointtype, so it'll become00000000000000000000000011100000.In step 4, when it is converted to
int, the most significant bit is0.Why, when I use the
%hdspecifier or even the%dspecifier, will it still still prompt me for a negative value instead of a positive?
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不，short 和 int 都是有符号类型，因此它是通过符号扩展而不是 0 字节填充来提升的：
将 MSB 保留为 1，即负数。
您可以通过先将其强制转换为无符号来伪造您所期望的行为，例如
No, short and int are both signed types, so it is promoted by sign extension not 0-byte padding:
leaving the MSB as 1 i.e. negative.
You could fake the behavour you're expecting by casting it unsigned first, e.g.
将short 转换为int 基本上是将short 的最高有效位复制到int 的前16 位中。这就是为什么 int 被打印为负数。如果您不希望出现此行为，请使用
ushort。Converting a short to an int basically replicates the most significate bit of the short into the top 16 bits of the int. This is why the int is printed as negative. If you do not want this behaviour using a
ushort.正如你所说，它是转换的，在这种情况下转换意味着知识。也就是说，编译器知道有符号短整型转换是如何工作的。它不只是在前面附加位，它还会创建一个与 Short 具有相同值的新 int。这就是为什么你会得到正确的数字。
As you say it is converted and conversion in this case implies knowlegde. That is the compiler knows how signed short to int conversion work. It does not just append bits in front, it creates a new int with the same value as the short. That's why you get the correct number.