Printf问题。

praeiji< sp ******* @ gmail.com>写道：

praeiji <sp*******@gmail.com> wrote:

unsigned char a = 200;
char b = 200;
printf（"％d％d"，a，b）;

给出：

200，-56

怎么来的？我没告诉printf第一个参数是未签名的
并且它自己检测到了它。 varargs似乎没有可能。
怎么可能？

unsigned char a = 200;
char b = 200;
printf( "%d %d", a, b );

gives :

200, -56

How comes? I didn''t tell printf that the first argument was unsigned
and it detected it on its own. It doesn''t seem possible with varargs.
How is it possible?

两个参数在传递给printf（）之前都被提升为int br />

_Ico


-

：wq

^ X ^ Cy ^ K ^ X ^ C ^ C ^ C ^ C

Both arguments get promoted to int before they are passed to printf()

_Ico

--
:wq
^X^Cy^K^X^C^C^C^C

谢谢，


但是他们为什么要对int进行类型转换呢？它是否来自gcc，当它看到

"％d"和printf函数？

Thanks,

But why are they typecasted to int? Does it come from gcc when it sees
"%d" and the printf function?

praeiji写道：

praeiji wrote:

unsigned char a = 200;
char b = 200;
printf（"％d％d"，a，b）;

给出：

200，-56

怎么来的？我没告诉printf第一个参数是未签名的
并且它自己检测到了它。 varargs似乎不太可能。
怎么可能？

谢谢，

Hi,

I was wondering something today. The following code :

unsigned char a = 200;
char b = 200;
printf( "%d %d", a, b );

gives :

200, -56

How comes? I didn''t tell printf that the first argument was unsigned
and it detected it on its own. It doesn''t seem possible with varargs.
How is it possible?

Thanks,

转换参数转换默认Pre -

Specifier Type Value Base cision

％d int x（int）x 10 1


这意味着％d格式为int 。如果你给它char和unsigned char它

将它们转换为int。

Conversion Argument Converted Default Pre-
Specifier Type Value Base cision
%d int x (int)x 10 1

That means %d formats int. If you give it char and unsigned char it
will convert them to int.