当我汇总初始化数组而gcc没有时，Clang警告我 [英] Clang warns me when I aggregate initialize an array while gcc doesn't

问题描述

当我使用CLANG编译以下代码片段时：

  #include< iostream> 
 #include< array> 
 #include< algorithm> 
 #include< functional> 
 
 int main（）{
 std :: array< int，2> a = {1,2}; 
 std :: array< int，2> b = {2,1}; 
 std :: array< int，2> C; 
 std :: transform（a.begin（），a.end（），b.begin（），c.begin（），std :: multiplies< int>（））; （auto&& i：c）std :: cout<<<我<< ; 
 std :: cout<<的std :: ENDL; 
} 
  

发出以下命令：

clang ++ -std = c ++ 14 -O2 -Wall -pedantic -pthread main.cpp

它发出警告：

警告：在子对象初始化
[-Wmissing-braces]

CLANG DEMO 然而，海湾合作委员会编写了这个程序，警告。

GCC DEMO

Q：

2. Clangs警告我的原因是什么？
在某些情况下，大括号可以被忽略。这是其中的一种情况。用于初始化 a 和 b 的最外括号是可选的。无论哪种方式，它在语法上都是正确的 - 但只是包含它们就更清楚了。 Clang只是警告你（警告，而不是错误） - 这是一个完全有效的警告。正如 chris 指出，使用 -Wmissing-braces ，gcc发出相同的警告。最终，两个编译器都接受代码，这是正确的;毕竟，这是一个有效的程序。这就是关键所在。




从[dcl.init.aggr]：






大括号可以省略在初始化列表中，如下所示。如果 initializer-list 以左大括号开头，则
后续的逗号分隔 初始化子句列表将初始化成员的一个子集;与成员相比，存在更多的初始化子句是
错误。但是，如果子集合
的 initializer-list 不以左大括号开头，则列表中只有足够的初始化子集会被带到
初始化子集的成员;任何剩余的初始化子句 剩下来初始化当前子集合所属的集合的下一个
成员。

  float y [4] [3] = {
 {1 ，3，5}，
 {2,4,6}，
 {3,5,7}，
}; 
  

是一个完全支撑的初始化： 1 ， 3 和 5 初始化数组的第一行 y [0] ，即 y [0] [0] ，
y [0] [1] 和 y [0] [2] 。同样，接下来的两行初始化 y [1] 和 y [2] 。初始化器提前结束，
，因此 y [3] s元素被初始化，就像使用 float（） ，
即用 0.0 进行初始化。在以下示例中，初始化程序列表中的花括号被省略;但是 initializer-list 的
与上例中的完全支撑的 initializer-list 具有相同的效果，

  float y [4] [3] = {
 1,3,5,2,4,6,3,5,7 
} ; 
  

y 的初始值设定项从左开始大括号，但是 y [0] 不是，所以使用了
列表中的三个元素。同样地，接下来的三个也被连续地用于 y [1] 和 y [2] 。 - 结束示例]







When I compile the following piece of code with CLANG:

#include <iostream>
#include <array>
#include <algorithm>
#include <functional>

int main() {
  std::array<int, 2> a = {1, 2};
  std::array<int, 2> b = {2, 1};
  std::array<int, 2> c;
  std::transform(a.begin(), a.end(), b.begin(), c.begin(), std::multiplies<int>());
  for(auto &&i : c) std::cout << i << " ";
  std::cout << std::endl;
}

by issuing the command:

clang++ -std=c++14 -O2 -Wall -pedantic -pthread main.cpp

It issues the warning:

warning: suggest braces around initialization of subobject [-Wmissing-braces]

CLANG DEMO

However, GCC compiles this program with out issuing a warning at all.

GCC DEMO

Q:

1. Which compiler is right?
2. What's the reason that Clangs warns me?

解决方案

In some cases, braces can be elided. This is one of those cases. The outer-most braces for initializing a and b are optional. It is syntactically correct either way - but it's clearer to just include them. Clang is just warning you (warning, not error) about this - it's a perfectly valid warning. And as chris, points out, with -Wmissing-braces, gcc issues the same warning. Ultimately, both compilers accept the code, which is correct; it is, after all, a valid program. That's all that matters.

From [dcl.init.aggr]:

Braces can be elided in an initializer-list as follows. If the initializer-list begins with a left brace, then the succeeding comma-separated list of initializer-clauses initializes the members of a subaggregate; it is erroneous for there to be more initializer-clauses than members. If, however, the initializer-list for a subaggregate does not begin with a left brace, then only enough initializer-clauses from the list are taken to initialize the members of the subaggregate; any remaining initializer-clauses are left to initialize the next member of the aggregate of which the current subaggregate is a member. [ Example:

float y[4][3] = {
    { 1, 3, 5 },
    { 2, 4, 6 },
    { 3, 5, 7 },
};

is a completely-braced initialization: 1, 3, and 5 initialize the first row of the array y[0], namely y[0][0], y[0][1], and y[0][2]. Likewise the next two lines initialize y[1] and y[2]. The initializer ends early and therefore y[3]s elements are initialized as if explicitly initialized with an expression of the form float(), that is, are initialized with 0.0. In the following example, braces in the initializer-list are elided; however the initializer-list has the same effect as the completely-braced initializer-list of the above example,

float y[4][3] = {
    1, 3, 5, 2, 4, 6, 3, 5, 7
};

The initializer for y begins with a left brace, but the one for y[0] does not, therefore three elements from the list are used. Likewise the next three are taken successively for y[1] and y[2]. —end example ]

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