如何使用＆QUOT; ZD＆QUOT;符用`的printf（）`？ [英] How to use "zd" specifier with `printf()`?

问题描述

寻找澄清关于使用ZD与的printf（）。

当然，以下是C99正确及更高版本。

 无效print_size（为size_t SZ）{
  的printf（％祖\\ n，SZ）;
}
 

的C规格的看起来的允许的printf（％ZD \\ n，SZ）取决于它是如何阅读：

7.21.6.1的 fprintf中函数

  

以Z 指定后续的 D ， I ， 0 ， U ， X 或 X 转换说明适用于为size_t 或相应的符号整型参数;或者说一个如下的 N 转换说明适用于指针对应的为size_t 参数的符号整型。 C11dr§7.21.6.17

这应该被理解为

1. 以Z 指定后续的 D ...转换说明适用于为size_t 或相应的符号整型参数...（两种）和以Z 指定后续的ū ...转换说明适用于为size_t 或相应的符号整型参数...（两种）

或

<醇开始=2>

以Z 指定后续的 D ...转换说明适用于相应的符号整型参数。 ..（符号类型只）和以Z 指定后续的 U ...转换说明适用于为size_t （无符号类型只）。

我已经使用了＃2的定义，但现在不那么肯定。

  

这是正确的，1，2，或其他什么东西？

  
  

如果＃2是正确的，什么是一个类型，可以使用的例子ZD％？

解决方案

的printf 与％ZD格式要求对应于无符号类型为size_t 的签署类型的参数。

标准C不为这种类型或一个很好的方式，以确定它是什么提供一个名称。如果为size_t 是无符号长，例如，那么％ZD的typedef 预计类型的参数长，但这不是便携式的假设。

该标准要求相应的符号和无符号类型用于该重新presentable这两种类型的非负值同样进行再presentation。一个脚注说，这是为了暗示他们是作为函数参数互换。因此，这：

 为size_t S = 42;
的printf（S =％ZD \\ N，S）;
 

应该工作，并应打印出42。它将间preT值 42 ，无符号的类型的为size_t ，就好像它是对应的符号的类型。但是，真的没有很好的理由这样做，因为％俎也是正确的，明确的，而不是诉诸额外的语言规则。和％俎工程的所有的类型的值为size_t ，包括外范围对应的符号类型。

最后，POSIX定义了一个标题类型 ssize_t供 <$c$c><unistd.h>和<一个href=\"http://pubs.opengroup.org/onlinepubs/9699919799/basedefs/sys_types.h.html\"><$c$c><sys/types.h>.虽然POSIX没有明确这么说，presumably ssize_t供将是对应于为size_t 有符号的类型。
所以，如果你正在编写特定POSIX-code，ZD％是（可能）类型的印刷值的正确格式 ssize_t供。

Looking for clarification on using "zd" with printf().

Certainly the following is correct with C99 and later.

void print_size(size_t sz) {
  printf("%zu\n", sz);
}

The C spec seems allow printf("%zd\n", sz) depending on on how it is read:

7.21.6.1 The fprintf function

z Specifies that a following d, i, o, u, x, or X conversion specifier applies to a size_t or the corresponding signed integer type argument; or that a following n conversion specifier applies to a pointer to a signed integer type corresponding to size_t argument. C11dr §7.21.6.1 7

Should this be read as

1. "z Specifies that a following d ... conversion specifier applies to a size_t or the corresponding signed integer type argument ... "(both types) and "z Specifies that a following u ... conversion specifier applies to a size_t or the corresponding signed integer type argument ..." (both types)

OR

2. "z Specifies that a following d ... conversion specifier applies to a corresponding signed integer type argument ..." (signed type only) and "z Specifies that a following u ... conversion specifier applies to a size_t" (unsigned type only).

I've been using the #2 definition, but now not so sure.

Which is correct, 1, 2, or something else?

If #2 is correct, what is an example of a type that can use "%zd"?

解决方案

printf with a "%zd" format expects an argument of the signed type that corresponds to the unsigned type size_t.

Standard C doesn't provide a name for this type or a good way to determine what it is. If size_t is a typedef for unsigned long, for example, then "%zd" expects an argument of type long, but that's not a portable assumption.

The standard requires that corresponding signed and unsigned types use the same representation for the non-negative values that are representable in both types. A footnote says that this is meant to imply that they're interchangeable as function arguments. So this:

size_t s = 42;
printf("s = %zd\n", s);

should work, and should print "42". It will interpret the value 42, of the unsigned type size_t, as if it were of the corresponding signed type. But there's really no good reason to do that, since "%zu" is also correct and well defined, without resorting to additional language rules. And "%zu" works for all values of type size_t, including those outside the range of the corresponding signed type.

Finally, POSIX defines a type ssize_t in the headers <unistd.h> and <sys/types.h>. Though POSIX doesn't explicitly say so, presumably ssize_t will be the signed type corresponding to size_t. So if you're writing POSIX-specific code, "%zd" is (probably) the correct format for printing values of type ssize_t.

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