这个constexpr整数不是一个空指针常量吗? [英] Is this constexpr integer not a null pointer constant?
问题描述
考虑下面的C ++ 11程序,其结果在GCC 4.7.2中:
int main b $ b {
constexpr int i = 0;
int * p = i;
}
// g ++ -g -ggdb -Wall -Wextra -pedantic -std = c ++ 11 t.cpp
// t.cpp:In function'int main()':
// t.cpp:4:13:错误:从'int'到'int *'无效转换[-fpermissive]
// t.cpp:4:警告:未使用的变量'p'[-Wunused-variable]
根据标准:
[C ++ 11:4.10 / 1]:
是一个整数常数表达式(5.19),整数类型的prvalue值为零[..]
我不能完全解析它,但我们不会期望 i
满足这个标准,并作为一个空指针常量因此不需要为 p
?
$ b的初始化显式转换
int *
$ b 如果I s / constexpr / const /
,编译成功,并使用 -ansi
code> -std = c + + 11 。
p>[C ++ 11:5.19 / 3]:
字面常量表达式是字面值类型的核心常量表达式,类型。 整数常数表达式是整数或无范围枚举类型的文字常量表达式。 [..]
和:
[C ++ 11:3.9 / 10]:
类型是字符类型 / p>
- 标量类型;或
- 引用类型;或
- 一个类类型(第9条),它具有以下所有属性:[..]
- li>
此时,我找不到该代码不符合SSL规定的原因,因此我怀疑有一个GCC错误。
然而,它可能是一个故意的错误,因为你引用4.10的段落被提议改变( active issue#903 ),因此这实际上是不符合SSL规定的代码。
如果I
s / constexpr / const /
,并使用-ansi
而不是-std = c ++ 11
/ p>
的定义明确允许在C ++ 03中使用这种情况:
[C ++ 03:5.19 / 1]:
[..]常量表达式(8.5),积分的非类型模板参数(8.5),非常量表达式的非类型模板参数(8.5),常数表达式(或枚举类型,以及sizeof
表达式。 [..]
Consider the following C++11 program, and its result in GCC 4.7.2:
int main()
{
constexpr int i = 0;
int* p = i;
}
// g++ -g -ggdb -Wall -Wextra -pedantic -std=c++11 t.cpp
// t.cpp: In function 'int main()':
// t.cpp:4:13: error: invalid conversion from 'int' to 'int*' [-fpermissive]
// t.cpp:4:9: warning: unused variable 'p' [-Wunused-variable]
According to the standard:
[C++11: 4.10/1]:
A null pointer constant is an integral constant expression (5.19) prvalue of integer type that evaluates to zero [..]
5.19 is a mess and I've failed to parse it fully, but wouldn't we expect i
to satisfy this criterion and act as a null pointer constant, consequently requiring no explicit conversion to int*
for the initialisation of p
?
The compilation succeeds if I s/constexpr/const/
and compile with -ansi
rather than -std=c++11
.
[C++11: 5.19/3]:
A literal constant expression is a prvalue core constant expression of literal type, but not pointer type. An integral constant expression is a literal constant expression of integral or unscoped enumeration type. [..]
And:
[C++11: 3.9/10]:
A type is a literal type if it is:
- a scalar type; or
- a reference type; or
- a class type (Clause 9) that has all of the following properties: [..]
- an array of literal type.
At this point, I can't find a reason for that code to be non-compliant, so I suspect a GCC bug.
However it may be a deliberate bug given that the passage you quoted out of 4.10 is proposed to be changed (active issue #903) so that this would in fact be non-compliant code.
The compilation succeeds if I
s/constexpr/const/
and compile with-ansi
rather than-std=c++11
.
The definition of integral constant expression explicitly allowed this case in C++03:
[C++03: 5.19/1]:
[..] An integral constant-expression can involve only literals (2.13), enumerators, const variables or static data members of integral or enumeration types initialized with constant expressions (8.5), non-type template parameters of integral or enumeration types, andsizeof
expressions. [..]
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