打印短而不是int ...？什么是“％hd”？ [英] Print a short instead of an int... ? What's up with "%hd"?
问题描述
将参数传递给VAL函数时，所有整数类型都是
提升，浮点数变为双精度数。因此，签名的短片将被提升为签署的int
。因此，我看不出你为什么要b $ b有任何理由：
printf（"％hd"，my_short）;
而不是：
printf（"％d"，my_short）;
Martin
When passing arguments to a VAL function, all integer types are
promoted, and floats become doubles. Therefore, signed short will be
promoted to signed int. Therefore I can''t see any reason why you''d
have:
printf( "%hd", my_short);
instead of:
printf( "%d", my_short);
Martin
推荐答案
Martin Wells写道：
Martin Wells wrote:
将参数传递给VAL函数时，所有整数类型
晋升，浮动成为双打。因此，签名的短片将被提升为签署的int
。因此，我看不出你为什么要b $ b有任何理由：
printf（"％hd"，my_short）;
而不是：
printf（"％d"，my_short）;
When passing arguments to a VAL function, all integer types are
promoted, and floats become doubles. Therefore, signed short will be
promoted to signed int. Therefore I can''t see any reason why you''d
have:
printf( "%hd", my_short);
instead of:
printf( "%d", my_short);
我认为目的是允许对printf（）和scanf（）使用相同的格式字符串
。 'h''在打印过程中不会增加任何内容，但会对扫描过程产生影响。
I believe that the purpose is to allow the use of the same format string
for printf() and scanf(). The ''h'' adds nothing during the printing
process, but makes a difference in the scanning process.
" James Kuyper Jr "写道：
"James Kuyper Jr." wrote:
>
Martin Wells写道：
>
Martin Wells wrote:
传递时VAL函数的参数，所有整数类型都是
提升，浮点数变为双精度数。因此，签名的短片将被提升为签署的int
。因此，我看不出你为什么要b $ b有任何理由：
printf（"％hd"，my_short）;
而不是：
printf（"％d"，my_short）;
When passing arguments to a VAL function, all integer types are
promoted, and floats become doubles. Therefore, signed short will be
promoted to signed int. Therefore I can''t see any reason why you''d
have:
printf( "%hd", my_short);
instead of:
printf( "%d", my_short);
我认为目的是允许对printf（）和scanf（）使用相同的格式字符串
。 'h''在打印过程中不会增加任何内容，但会对扫描过程产生影响。
I believe that the purpose is to allow the use of the same format string
for printf() and scanf(). The ''h'' adds nothing during the printing
process, but makes a difference in the scanning process.
它可能有所作为。考虑％hx当传递（短）-1，
在我的系统上打印ffff，而％x打印ffffffff。
-
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不要 - 邮寄给我：< mailto：Th ************* @ gmail.com>
It could make a difference. Consider "%hx" when passed (short)-1,
which on my system prints "ffff", whereas "%x" prints "ffffffff".
--
+-------------------------+--------------------+-----------------------+
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Don''t e-mail me at: <mailto:Th*************@gmail.com>
Kenneth：
Kenneth:
它可能会有所作为。考虑％hx当传递（短）-1，
在我的系统上打印ffff，而％x打印ffffffff。
It could make a difference. Consider "%hx" when passed (short)-1,
which on my system prints "ffff", whereas "%x" prints "ffffffff".
好的我可以看到以下情况如何：
（无符号）（短）-1 == UINT_MAX（也许是0xffff）
（短无符号）（短）-1 == USHRT_MAX（也许是0xffffffff）
....但它不是未定义的行为如果需要无符号整数类型，请提供带有签名
整数类型的printf，反之亦然？
如果我想打印无符号的最大值简短（当然不使用
USHRT_MAX），然后我会选择：
printf（"％u"，（unsigned）（short）未签名的）-1）
马丁
OK I can see how the following are true:
(unsigned)(short)-1 == UINT_MAX (maybe 0xffff)
(short unsigned)(short)-1 == USHRT_MAX (maybe 0xffffffff)
....but is it not undefined behaviour to supply printf with a signed
integer type when it expects an unsigned integer type, or vice versa?
If I wanted to print the max value of an unsigned short (without using
USHRT_MAX of course), then I''d go with:
printf("%u",(unsigned)(short unsigned)-1)
Martin
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