为什么用逗号进行双重初始化是非法的？
我有三个代码片段。这个：
1,7; //yes, that's all the code
编译没问题。这个：
double d = (1, 7);
也可以编译。然而这个：
double d = 1, 7;
无法编译。 gcc-4.3.4 说
错误：数字常量之前应有非限定 ID
，Visual C++ 10 表示
错误 C2059：语法错误：“常量”
为什么会有这样的差异？为什么这三个都不用 编译， 在这三个中具有相同的效果？
I have three code snippets. This one:
1,7; //yes, that's all the code
compiles okay. This one:
double d = (1, 7);
also compiles okay. Yet this one:
double d = 1, 7;
fails to compile. gcc-4.3.4 says
error: expected unqualified-id before numeric constant
and Visual C++ 10 says
error C2059: syntax error : 'constant'
Why such difference? Why don't all the three compile with , having the same effect in all three?
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在前两种情况下，语句使用 C++ 的 逗号运算符
在后一种情况下，逗号被用作单独的变量，编译器希望您声明多个标识符；此处不使用逗号作为运算符。
最后一种情况类似于：
当您执行此操作时：
编译器需要变量标识符而不是数字常量。因此 7 在这里是非法的。
然而，当你这样做时：
正在使用普通的逗号运算符：1 被评估并丢弃，而 7 存储在 d 中。
In the first two cases, the statements are using C++'s comma operator
In the latter case, comma is being used as variable separate and the compiler is expecting you to declare multiple identifiers; the comma is not being used as the operator here.
The last case is similar to something like:
When you do this:
The compiler expects a variable identifier and not a numeric constant. Hence 7 is illegal here.
However when you do this:
the normal comma operator is being used: 1 gets evaluated and discard while 7 is stored in d.
区别在于，在
1, 7;和(1, 7)中，表达式中允许使用逗号运算符。最后一个示例
是一个声明，其中逗号不是运算符而是分隔符。编译器期望类似这样的内容
，这将是正确的变量声明。
请注意，逗号有时是一个运算符（在表达式中），但也可以在其他地方用作分隔符，例如函数声明中的参数列表。
The difference is that in
1, 7;and(1, 7)you have expressions where a comma operator is allowed.Your last example
is a declaration, where the comma isn't an operator but a separator. The compiler exepcts something like
which would be a correct variable declaration.
Note that the comma is sometimes an operator (in expressions), but is also used as a separator in other places like parameter lists in function declarations.
double d = (1, 7);这里将评估(1, 7)第一的;逗号用作顺序求值运算符，并且
7将被分配给d。double d = 1, 7;在这种情况下有一个问题：该部分逗号之前意味着您声明一个 double 并设置其值，但是
逗号后面的部分是无意义，因为它只是一个
整数常量。
double d = (1, 7);Here the(1, 7)will be evaluatedfirst; the comma works as sequential-evaluation operator, and
7will be assigned tod.double d = 1, 7;In this case there is a problem: the partbefore the comma means you declare a double and set its value, but
the part after the comma is meaningless, because it's just a single
integer constant.
我相信这是因为最后一个被视为（不正确的）声明行：
(double d = 1), (7)I believe it is because the last one is treated as (incorrect) declaration line:
(double d = 1), (7)