15  

That leeway wording doesn't mean that any and everything in C99 is in C++11. What you quoted is just introductory text.

这种灵活的措辞并不意味着C99中的所有内容都是在c++ 11中。你所引用的只是引言。

13  

This C99 feature is effectively overridden by C++'s own semantics, as can be any otherwise "inherited" feature:

这个C99特性被c++自身的语义所覆盖，可以是任何其他“继承”特性:

[C++11: 8.3.4/1]: In a declaration T D where D has the form

[C+ 11: 8.3.4/1]:在声明T D中，D有表格

D1 [ constant-expression_opt ] attribute-specifier-seq_opt

D1 [constant-expressionopt]属性-specifier-seqopt

[..]

[. .]

This is the only array declaration syntax we're given in C++.

这是c++中唯一给出的数组声明语法。

Note that no mention of this difference is given in the "compatibility with C" clause C.1.

注意，在“与C兼容”条款C.1中没有提到这种差异。

7  

The definition of constant-expression is different for the two languages.

对于两种语言来说，常量表达式的定义是不同的。

const size_t size = 5;
int array[size]; // array in C++, VLA in C

2  

This compiles for me: (g++ 4.6 with -std=c++0x). But it doesn't compile with -pedantic (thanks @MarkB). Instead it warns that "template.cpp:7:12: warning: ISO C++ forbids variable length array ‘n’ [-Wvla]"

这为我编译:(g+ 4.6与-std=c++0x)。但是它没有使用-pedantic编译(谢谢@MarkB)。相反，它警告“模板”。警告:ISO c++禁止可变长度数组' n ' [-Wvla]"

int main(int argc, char ** argv) {
    int n[argc];
}

So the size of n can not be known at compile time by the compiler. Is this a GNU extension to C++? This does appear to be a GNU extension, and that VLAs are not an official part of C++11.

所以编译器在编译时不能知道n的大小。这是c++的GNU扩展吗?这似乎是一个GNU扩展，并且VLAs不是c++ 11的官方部分。

(Of course, I'm just playing with a compiler. So take this with a pinch of salt.)

当然，我只是在玩编译器。所以，我们还是少吃一点吧。