问题
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?
回答1:
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:
float x,y;
float a = 10, b = 20;
When you do this:
double d = 1, 7;
The compiler expects a variable identifier and not a numeric constant. Hence 7 is illegal here.
However when you do this:
double d = (1,7);
the normal comma operator is being used: 1 gets evaluated and discard while 7 is stored in d.
回答2:
The difference is that in 1, 7; and (1, 7) you have expressions where a comma operator is allowed.
Your last example
double d = 1, 7;
is a declaration, where the comma isn't an operator but a separator. The compiler exepcts something like
double d = 1, e = 7;
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.
回答3:
double d = (1, 7);Here the(1, 7)will be evaluated first; the comma works as sequential-evaluation operator, and7will be assigned tod.double d = 1, 7;In this case there is a problem: the part before 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.
回答4:
I believe it is because the last one is treated as (incorrect) declaration line: (double d = 1), (7)
来源：https://stackoverflow.com/questions/8519884/why-is-this-double-initialization-with-a-comma-illegal