为什么用逗号双重初始化是非法的? [英] Why is this double initialization with a comma illegal?

问题描述

我有三个代码段.这个:

I have three code snippets. This one:

1,7; //yes, that's all the code

编译正常.这个:

double d = (1, 7);

也可以编译.然而这个:

also compiles okay. Yet this one:

double d = 1, 7;

无法编译.gcc-4.3.4说

fails to compile. gcc-4.3.4 says

错误:数字常量前应有预期的unqualified-id

error: expected unqualified-id before numeric constant

Visual C ++ 10说

and Visual C++ 10 says

错误C2059:语法错误:恒定"

error C2059: syntax error : 'constant'

为什么有这种区别?为什么这三个都不能用 进行编译，而在所有三个中都具有相同的效果?

Why such difference? Why don't all the three compile with , having the same effect in all three?

推荐答案

在前两种情况下，语句使用C ++的

In the first two cases, the statements are using C++'s comma operator

在后一种情况下，逗号被用作单独的变量，并且编译器期望您声明多个标识符；逗号不在此处用作运算符.

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.

最后一种情况类似于:

float x,y;
float a = 10, b = 20;

执行此操作时:

double d = 1, 7;

编译器期望变量标识符而不是数字常数.因此，这里7是非法的.

The compiler expects a variable identifier and not a numeric constant. Hence 7 is illegal here.

但是，当您这样做时:

double d = (1,7);

使用了普通的逗号运算符:1被求值并丢弃，而7存储在d中.

the normal comma operator is being used: 1 gets evaluated and discard while 7 is stored in d.

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