stdint类型和转换...... [英] stdint types and conversion...
问题描述
大家好，
INT8_C，INT16_C等名称的转换宏在
stdint.h中定义，将其参数转换为合适的表示形式对于
它们对应的类型，即int8_t，int16_t等。
我的问题是：
1.转换宏应该是甚至在从相同类型的
变量中分配时使用？
2.为什么没有为int_fast8_t等提供这样的宏以及
int_least8_t等？可以为他们分配其他类型的值吗？
没有大小不匹配并期望发生正确的事情？
谢谢。
Hello all,
Conversion macros along the name of INT8_C, INT16_C etc, are defined in
stdint.h to convert their argument into suitable representations for
their corresponding types, i.e. int8_t, int16_t etc.
My questions are:
1. Should the conversion macros be used even when assigning from a
variable of the same type?
2. Why are no such macros provided for int_fast8_t etc. as well as
int_least8_t etc? Can they be assigned values from other types provided
there are no size mismatches and expect the correct thing to happen?
Thanks.
推荐答案
santosh写道：
santosh wrote:
大家好，
转换宏名为INT8_C，INT16_C等等，在stdint.h中定义，将它们的参数转换为适当的表示形式，用于它们对应的类型，即int8_t，int16_t等。
这些不是* c * onversion宏，它们用于声明适当类型的* c * onstants
。 "转换"暗示某些东西正在被转换，
但事实并非如此;常量将具有适当的类型。您的
其他问题也来自于这种误解。
我的问题是：
1.即使从
变量中分配转换宏也应该使用相同类型的？
不可以。事实上，你必须*不*使用它们：
"这些宏的任何实例中的参数都应该是十进制，八进制或
十六进制常量（如6.4.4.1中所定义），其值不是
超过相应类型的限制。 [7.8.14，第2段]。
任何类型的变量都不是（必须是）常数。
2.为什么不这样为int_fast8_t等提供的宏以及
int_least8_t等？可以为他们分配其他类型的值吗？
没有大小不匹配并期望发生正确的事情？
Hello all,
Conversion macros along the name of INT8_C, INT16_C etc, are defined in
stdint.h to convert their argument into suitable representations for
their corresponding types, i.e. int8_t, int16_t etc.
These are not *c*onversion macros, they are used to declare *c*onstants of
the appropriate types. "Conversion" implies something is being converted,
which is not the case; the constant will have the appropriate type. Your
other questions follow from this misconception.
My questions are:
1. Should the conversion macros be used even when assigning from a
variable of the same type?
No. In fact, you must *not* use them:
"The argument in any instance of these macros shall be a decimal, octal, or
hexadecimal constant (as defined in 6.4.4.1) with a value that does not
exceed the limits for the corresponding type." [7.8.14, paragraph 2].
A variable of any kind is not (required to be) a constant.
2. Why are no such macros provided for int_fast8_t etc. as well as
int_least8_t etc? Can they be assigned values from other types provided
there are no size mismatches and expect the correct thing to happen?
编号通常的规则适用：你需要一个演员表价值可能不合适，
并且你有责任确保它们合适。 （一个粗略的过度简化
的实际规则，当然，看看有关C的隐含和
显式转换的任何好书。）
S.
No. The usual rules apply: you need a cast for values that might not fit,
and it''s your responsibility to ensure they fit. (A gross oversimplification
of the actual rules, of course, see any good book on C about implicit and
explicit conversion.)
S.
Skarmander写道：
Skarmander wrote:
santosh写道：
santosh wrote:
大家好，
INT8_C，INT16_C等名称的转换宏在
stdint.h中定义，将其参数转换为适当的表示形式，即相应的类型，即int8_t，int16_t等。
Hello all,
Conversion macros along the name of INT8_C, INT16_C etc, are defined in
stdint.h to convert their argument into suitable representations for
their corresponding types, i.e. int8_t, int16_t etc.
这些不是* c * onversion宏，它们用于声明相应类型的* c * onstants。 "转换"暗示某些东西正在转变，
并非如此;常量将具有适当的类型。您的其他问题也来自于这种误解。
These are not *c*onversion macros, they are used to declare *c*onstants of
the appropriate types. "Conversion" implies something is being converted,
which is not the case; the constant will have the appropriate type. Your
other questions follow from this misconception.
如果我理解正确的话，在分配时将使用宏
字面常量它们各自类型的变量值？
类似于：
....
int8_t i8 = INT8_C（-112 ）;
....
If I understand correctly the macros are to be used when assigning
literal constant values to variables of their respective types?
Something like:
....
int8_t i8 = INT8_C(-112);
....
我的问题是：
1.是否应使用转换宏即使从相同类型的变量中分配？
My questions are:
1. Should the conversion macros be used even when assigning from a
variable of the same type?
否。实际上，你必须*不*使用它们：
任何参数这些宏的实例应为十进制，八进制或
十六进制常量（如6.4.4.1中所定义），其值不会超过相应类型的限制。 [7.8.14，第2段]。
任何类型的变量都不是（必须是）常数。
No. In fact, you must *not* use them:
"The argument in any instance of these macros shall be a decimal, octal, or
hexadecimal constant (as defined in 6.4.4.1) with a value that does not
exceed the limits for the corresponding type." [7.8.14, paragraph 2].
A variable of any kind is not (required to be) a constant.
2.为什么没有这样的为int_fast8_t等提供的宏以及
int_least8_t等？是否可以为其他类型的值分配
没有大小不匹配并且期望发生正确的事情？
2. Why are no such macros provided for int_fast8_t etc. as well as
int_least8_t etc? Can they be assigned values from other types provided
there are no size mismatches and expect the correct thing to happen?
不适用通常的规则适用：您需要对可能没有的值进行强制转换合适，
确保它们合适是你的责任。 （实际规则的粗略过度简化，当然，请参阅关于隐式和明确转换的关于C的任何好书。）
S。
No. The usual rules apply: you need a cast for values that might not fit,
and it''s your responsibility to ensure they fit. (A gross oversimplification
of the actual rules, of course, see any good book on C about implicit and
explicit conversion.)
S.
感谢您的澄清。
Thanks for the clarification.
2006年3月25日04:30:08 -0800，santosh < sa ********* @ gmail.com>
写道：
On 25 Mar 2006 04:30:08 -0800, "santosh" <sa*********@gmail.com>
wrote:
大家好，
在stdint.h中定义了名为INT8_C，INT16_C等的转换宏，以将它们的参数转换为适当的表示形式，即
对应的类型，即int8_t，int16_t等。
我的n1124副本没有显示任何此类宏。你在哪里找到了
吗？他们为什么需要他们？您可以使用简单的
int8_t x;
定义所需
类型的变量，并且可以将表达式转换为所需类型简单
cast
x =（int8_t）< expression>
我的问题是：
1。即使从相同类型的变量分配时，是否应该使用转换宏？
2。为什么没有为int_fast8_t等提供这样的宏以及
int_least8_t等？可以为他们分配其他类型的值吗？
没有大小不匹配并期望正确的事情发生？
谢谢。
Hello all,
Conversion macros along the name of INT8_C, INT16_C etc, are defined in
stdint.h to convert their argument into suitable representations for
their corresponding types, i.e. int8_t, int16_t etc.
My copy of n1124 does not show any such macros. Where did you find
them? Why would they be needed? You define a variable of the desired
type with a simple
int8_t x;
and you can convert an expression to the desired type with a simple
cast
x = (int8_t)<expression>
My questions are:
1. Should the conversion macros be used even when assigning from a
variable of the same type?
2. Why are no such macros provided for int_fast8_t etc. as well as
int_least8_t etc? Can they be assigned values from other types provided
there are no size mismatches and expect the correct thing to happen?
Thanks.
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