printf（）一个size_t实例的便携方式 [英] Portable way to printf() a size_t instance
问题描述
您好，
我的代码可以使用size_t类型的变量。代码是
最初是在32位机器上开发的，但现在它在32位机器和64位机器上运行。
在代码中有这样的语句：
size_t buffer_size;
printf（"总缓冲区大小：％ud bytes \ n" ;，buffer_size）;
现在格式为％ud在32位计算机上运行良好;它实际上
也适用于64位计算机，但编译器吐出
警告消息。在64位计算机上，它首选：
printf（总缓冲区大小：％uld bytes \ n，buffer_size）; / *或者udl？ * /
正如我所说它有效，但我真的更喜欢这个程序在没有警告的情况下编译
，因为现在我得到*很多*警告类型
file.c：line：warning：unsigned int format，不同类型arg（arg 1）。
然后，打开有几次，这实际上让我错过了更多
重要警告。
任何提示赞赏。
Joakim
-
Joakim Hove
hove AT ift uib no /
Tlf：+47（55 5）8 27 90 / Stabburveien 18
传真：+47（55 5）8 94 40 / N-5231 Paradis
http://www.ift.uib.no/~hove/ / 55 91 28 18/92 68 57 04
Hello,
I have code which makses use of variables of type size_t. The code is
originally developed on a 32 bit machine, but now it is run on both a
32 bit and a 64 bit machine.
In the code have statements like this:
size_t buffer_size;
printf("Total buffer size: %ud bytes \n",buffer_size);
Now the format "%ud" works nicely on the 32 bit computer; it actually
works on the 64 bit computer as well, but the compiler spits out
warning message. On the 64 bit computer it would have prefered:
printf("Total buffer size: %uld bytes \n",buffer_size); /* Or udl? */
As I said it works, but I would really prefer the program to compile
without warnings, as it is now I get *many* warnings of the type
file.c:line: warning: unsigned int format, different type arg (arg 1).
And, on several occasion this has actually led me to miss more
important warnings.
Any tips appreciated.
Joakim
--
Joakim Hove
hove AT ift uib no /
Tlf: +47 (55 5)8 27 90 / Stabburveien 18
Fax: +47 (55 5)8 94 40 / N-5231 Paradis
http://www.ift.uib.no/~hove/ / 55 91 28 18 / 92 68 57 04
推荐答案
Joakim Hove写道：
Joakim Hove wrote:
你好，
我有代码可以使用size_t类型的变量。代码最初是在32位机器上开发的，但现在它可以在32位和64位机器上运行。
在代码中有这样的语句：
size_t buffer_size;
printf（"总缓冲区大小：％ud bytes \ n"，buffer_size）;
现在格式为％ud ;在32位计算机上运行良好;它实际上也适用于64位计算机，但编译器会发出警告消息。在64位计算机上，它首选：
printf（总缓冲区大小：％uld bytes \ n，buffer_size）; / *或者udl？ * /
正如我所说它有效，但我真的更喜欢这个程序在没有警告的情况下编译
，因为现在我得到了*很多*类型的警告
file.c：line：warning：unsigned int format，不同类型的arg（arg 1）。
而且，有几次这实际上让我错过了更多
重要的警告。
任何提示赞赏。
Hello,
I have code which makses use of variables of type size_t. The code is
originally developed on a 32 bit machine, but now it is run on both a
32 bit and a 64 bit machine.
In the code have statements like this:
size_t buffer_size;
printf("Total buffer size: %ud bytes \n",buffer_size);
Now the format "%ud" works nicely on the 32 bit computer; it actually
works on the 64 bit computer as well, but the compiler spits out
warning message. On the 64 bit computer it would have prefered:
printf("Total buffer size: %uld bytes \n",buffer_size); /* Or udl? */
As I said it works, but I would really prefer the program to compile
without warnings, as it is now I get *many* warnings of the type
file.c:line: warning: unsigned int format, different type arg (arg 1).
And, on several occasion this has actually led me to miss more
important warnings.
Any tips appreciated.
对于C89，转换为long unsigned。
printf （" sizeof（int）是％lu \ n"，（long unsigned）sizeof（int））;
-
pete
For C89, convert to long unsigned.
printf("sizeof(int) is %lu\n", (long unsigned)sizeof(int));
--
pete
Joakim Hove写道：
Joakim Hove wrote:
你好，
我有代码可以使用size_t类型的变量。代码最初是在32位机器上开发的，但现在它可以在32位和64位机器上运行。
在代码中有这样的语句：
size_t buffer_size;
printf（"总缓冲区大小：％ud bytes \ n"，buffer_size）;
现在格式为％ud ;在32位计算机上运行良好;
不应该。 ％< qualifier> d是签名
整数的转换说明符; ％< qualifier> u是unsigned
整数的转换说明符。 ''你'不是％d的限定符。
它实际上也适用于64位计算机，但编译器吐出
警告信息。
类似％ud并不代表任何东西？
在64位计算机上它会更喜欢：
printf（" Total buffer size：％uld bytes \ n"，buffer_size）; / *或者udl？ * /
Hello,
I have code which makses use of variables of type size_t. The code is
originally developed on a 32 bit machine, but now it is run on both a
32 bit and a 64 bit machine.
In the code have statements like this:
size_t buffer_size;
printf("Total buffer size: %ud bytes \n",buffer_size);
Now the format "%ud" works nicely on the 32 bit computer;
It shouldn''t. The %<qualifier>d are the conversion specifiers for signed
integers; the %<qualifier>u are the conversion specifiers for unsigned
integers. ''u'' is not a qualifier for %d.
it actually
works on the 64 bit computer as well, but the compiler spits out
warning message.
Something like "%ud doesn''t mean anything"?
On the 64 bit computer it would have prefered:
printf("Total buffer size: %uld bytes \n",buffer_size); /* Or udl? */
我对此表示怀疑。 printf例程很少想要没有意义的说明符。
如果你有一个最新的库，你可以使用
printf（总缓冲区大小：％zu bytes \ n，buffer_size ）;
否则，请尝试
printf（总缓冲区大小：％lu bytes \ n，
（ long unsigned）buffer_size）;
或
printf（"总缓冲区大小：％llu bytes \ n"，
（long long） unsigned）buffer_size）;
也就是说，使用（正确编写的）无符号说明符并将
size_t参数转换为相同大小的unsigned。
I doubt it. The printf routine rarely "wants" meaningless specifiers.
If you have an up-to-date library, you could use
printf("Total buffer size: %zu bytes \n",buffer_size);
Otherwise, try
printf("Total buffer size: %lu bytes \n",
(long unsigned)buffer_size);
or
printf("Total buffer size: %llu bytes \n",
(long long unsigned) buffer_size);
That is, use an (properly written) unsigned specifier and cast the
size_t argument to the same size unsigned.
Martin Ambuhl写道（现在澄清）：
Martin Ambuhl wrote (and now clarifies):
Joakim Hove写道：
[...]
Joakim Hove wrote: [...]
现在格式为％ud在32位计算机上运行良好;
Now the format "%ud" works nicely on the 32 bit computer;
它不应该。 ％< qualifier> d是签名
整数的转换说明符; ％< qualifier> u是无符号
整数的转换说明符。 ''你'不是％d的限定符。
It shouldn''t. The %<qualifier>d are the conversion specifiers for signed
integers; the %<qualifier>u are the conversion specifiers for unsigned
integers. ''u'' is not a qualifier for %d.
我应该更清楚。 "％U"是您正在使用的说明符。
" d"是输出字符串的文字部分。请参阅下面的示例
并注意字符串末尾的额外d：
#include< stdio.h>
int main（无效）
{
printf（"使用%% ud打印无符号值：％ud \ n，5u） ;
返回0;
}
使用％ud打印无符号值：5d
I should have been clearer. "%u" is the specifier you are using. The
"d" is a literal portion of the output string. See the example below
and note the extra ''d'' at the end of the string:
#include <stdio.h>
int main(void)
{
printf("Printing an unsigned value using %%ud: %ud\n", 5u);
return 0;
}
Printing an unsigned value using %ud: 5d
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