用浮点文本初始化const int [英] Initializing a const int with a floating point literal

问题描述

i = 1.0; //注意我将一个double赋值给这里的
char a [i];
}

问题

使用 g ++ -O0 -Wall -pedantic -ansi -std = c ++ 11 编译上述代码不会产生错误（除了未使用的变量）。但是，如果我删除 -std = c ++ 11 ，我会收到以下警告：

警告：ISO C ++禁止可变长度数组

根据这个问题，我相信在C ++ 03，代码是无效的。但是，有人可以解释一下规则在C ++ 11中的变化吗？

（这个问题是上一个问题我回答。）

解决方案

数组绑定必须是一个整数常数表达式，见8.3.4 [dcl.array] / 1（C ++ 03中的相同措辞， C ++ 11）：

如果常数表达式（5.19）存在，它将是一个整数常数表达式，

在C ++ 03中，一个整数常量表达式不能由浮点文本初始化，除非转换为整数类型，请参阅5.19 [expr.const] / 1的最后一句：

只有文字（2.13），枚举器， const 变量或使用常量表达式初始化的积分或枚举类型的静态数据成员（8.5），积分或枚举类型的非类型模板参数，和 sizeof 表达式。浮动文本（2.13.3）只有在转换为整数或枚举类型时才会出现。

这意味着在C ++ 03 i 不是一个整数常数表达式，因此不能用作数组边界。

GCC和Clang allow可变长度数组作为C ++ 03的扩展，所以它编译为非常量绑定，但是你会收到 -pedantic 的警告。更改常量的初始值以将其转换为整数类型使 i 是一个有效的整数常量表达式：

  const int i =（int）1.0; 
  

使用该更改，数组不再是变量长度，即使使用 -pedantic 。

在C ++ 11中5.19 [expr.const] / 3说：

字符常量表达式是字面类型的核心常量表达式，但不是指针类型。

前面的（完全的）常量表达式冗长的）段落描述核心常量表达式的规则，但基本上在C ++ 11中，双初始化器不会阻止 i 是核心常量表达式，即使没有强制转换，它是一个整数常数表达式，因此有效的数组界限，所以没有警告。

Example

int main()
{
    const int i = 1.0; // Notice I am assigning a double to an int here
    char a[i];
}

Question

Compiling the above code with g++ -O0 -Wall -pedantic -ansi -std=c++11 gives no errors (except for an unused variable). However, if I remove -std=c++11, I get the following warning:

warning: ISO C++ forbids variable length array

According to this SO question, I believe that in C++03, the code is invalid. However, can someone explain how the rule has changed in C++11?

(This question was a result of a previous question I answered.)

解决方案

An array bound must be an integral constant expression, see 8.3.4 [dcl.array]/1 (same wording in C++03 and C++11):

If the constant-expression (5.19) is present, it shall be an integral constant expression and its value shall be greater than zero.

In C++03 an integral constant expression cannot be initialized by a floating literal unless cast to integral type, see the last sentence of 5.19 [expr.const]/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, and sizeof expressions. Floating literals (2.13.3) can appear only if they are cast to integral or enumeration types.

This means that in C++03 i is not an integral constant expression, so cannot be used as an array bound.

GCC and Clang allow variable-length arrays as an extension to C++03, so it compiles with a non-constant bound, but you get a warning with -pedantic. Changing the constant's initializer to cast it to integral type makes i a valid integral constant expression:

const int i = (int) 1.0;

With that change the array is no longer variable length and there is no warning even with -pedantic.

In C++11 5.19 [expr.const]/3 says:

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.

The preceding (quite lengthy) paragraphs describe the rules for core constant expressions, but basically in C++11 the double initializer does not prevent i being a core constant expression, even without a cast, so it is an integral constant expression and therefore a valid array bound, so no warning.

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