初始化程序变量列表 [英] Initializer list of variables

问题描述

可以创建一个 initializer_list 的变量，例如函数参数（cf. function test ） ？下面的代码是可行的，既没有clang也没有gcc抱怨什么，但我只是想确保这是确定。

Is it possible to create an initializer_list of variables, like function arguments for example (cf. function test)? The code below works, and neither clang nor gcc complain about anything, but I would just like to make sure this is ok.

edit：将 wrap 的参数改为const ref

edit: changing the argument of wrap to a const ref

#include <iostream>
#include <initializer_list>

template <class T>
struct array
{
    T     *ptr;
    size_t len;

    array() { clear(); }
    array( T *p, size_t l ) { assign(p,l); }

    inline void clear() { ptr=nullptr; len=0; }
    inline void assign( T *p, size_t l ) { ptr=p; len=l; }

    inline T& operator[] ( size_t i ) const { return ptr[i]; }
};

template <class T>
inline array<const T> wrap( const std::initializer_list<T>& lst )
    { return array<const T>( lst.begin(), lst.size() ); }

void test( int a, int b, int c )
{ 
    auto ar = wrap({a,b,c}); 
    std::cout<< ar[2] << std::endl;
}

int main()
{
    auto a = wrap({1,2,3});
    std::cout<< a[2] << std::endl;

    test(1,2,3);
}

如果我试图在 test 中返回我的包数组，初始化列表 {a，b，c}

Side question; if I tried to return my wrapped array in test, the initializer list {a,b,c} would get out of scope, and the array I'm returning would be invalid -- is that correct?

推荐答案

 auto ar = wrap({a,b,c}); 

这将创建 int [3] ，然后将 initializer_list< int> 绑定到该数组，然后调用 wrap $ c> array< const int>

This create a temporary array of type int[3], then binds an initializer_list<int> to that array, then calls wrap which creates an array<const int> that refers to the array.

在表达式的末尾，数组被销毁， code> array< const int> 使用悬空指针，因此这是未定义的行为：

At the end of the expression the array is destroyed, leaving the array<const int> with a dangling pointer, so this is undefined behaviour:

 std::cout<< ar[2] << std::endl;

这也适用于 main ，变量 a 包含悬空指针， a [2] 是未定义的行为。

This also applies to the code in main, the variable a contains a dangling pointer and a[2] is undefined behaviour.

你可以通过用分配内存的类型数组替换 int 的数组来验证这一点，以便valgrind或asan注意到错误：

You can verify this by replacing the array of int with an array of types that allocate memory, so that valgrind or asan will notice the bug:

using V = std::vector<int>;
auto a = wrap({V{1}, V{2}, V{3}});
std::cout<< a[2].front() << std::endl;

现在 a [2] code> std :: vector< int> 对象，但试图访问其 front（）成员导致程序中止： / p>

Now a[2] is a std::vector<int> object, but trying to access its front() member causes the program to abort:

==28356==ERROR: AddressSanitizer: heap-use-after-free on address 0x60200000efb0 at pc 0x000000401205 bp 0x7fffa46f2900 sp 0x7fffa46f28f8
READ of size 4 at 0x60200000efb0 thread T0
    #0 0x401204 in main /tmp/il.cc:28
    #1 0x3236e21d64 in __libc_start_main (/lib64/libc.so.6+0x3236e21d64)
    #2 0x400ec8  (/tmp/a.out+0x400ec8)
...

或使用valgrind：

Or with valgrind:

==28364== Invalid read of size 4
==28364==    at 0x400C72: main (il.cc:28)
==28364==  Address 0x51dfd20 is 0 bytes inside a block of size 4 free'd
==28364==    at 0x4A07991: operator delete(void*) (vg_replace_malloc.c:502)
==28364==    by 0x4013BF: __gnu_cxx::new_allocator<int>::deallocate(int*, unsigned long) (new_allocator.h:110)
==28364==    by 0x4012F8: std::allocator_traits<std::allocator<int> >::deallocate(std::allocator<int>&, int*, unsigned long) (alloc_traits.h:386)
==28364==    by 0x4011B1: std::_Vector_base<int, std::allocator<int> >::_M_deallocate(int*, unsigned long) (stl_vector.h:178)
==28364==    by 0x40102A: std::_Vector_base<int, std::allocator<int> >::~_Vector_base() (stl_vector.h:160)
==28364==    by 0x400EC4: std::vector<int, std::allocator<int> >::~vector() (stl_vector.h:425)
==28364==    by 0x400C2A: main (il.cc:27)

侧问题;如果我试图在测试中返回我的包数组，初始化列表 {a，b，c} 会超出范围，我返回的数组将无效 - 是否正确？

Side question; if I tried to return my wrapped array in test, the initializer list {a,b,c} would get out of scope, and the array I'm returning would be invalid -- is that correct?

已经超出范围， ar 无效，甚至在您退回之前。

It's already out of scope and ar already invalid even before you return it.

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