算术运算是否无效？

Question

I know that the following, if possible, would be an absolutely bad practice, but I want to know if this is possible.

The question is the following: is it possible in C++ (and in a way the compiler does not throw any warning), to perform a useless arithmetic operation with a function returning a void.

std::vector<int> v;
int i = 42 + v.resize(42); 
/* How to transform the last line to execute resize and to have i = 42 */

I know that this is stupid, but that is not the question...

Answer 1

我不确定这是否有意义，但是您可以在此处使用逗号运算符 ：

int i = (v.resize(42), 42);

Answer 2

You could use the comma operator :

int i = (v.resize(42), 42);

and with GCC you could use its statement expression extension:

int i = ({v.resize(42); 42;})

and in standard C++11 you could use and call an anonymous closure :

int i = ([&v]() {v.resize(42); return 42;}());

Answer 3

Type void has no values so it may not be used in arithmetic expressions.

In my opinion the design of member function resize is bad. Instead of void it should return the object itself. In this case you could write for example

int i = v.resize(42).size(); 

I pointed out about this in the forum where the C++ Standard is discussed.

As for your question then you can write

int i = ( v.resize(42), v.size() );

using the comma operator.

Or maybe it would be better to separate these two calls

v.resize(42);
int i = v.size();

Answer 4

Don't see the point, but here's another way

std::tie(i, std::ignore) = std::make_tuple(42, (v.resize(42),1) );

Also you can do:

if ((i=42)) v.resize(42); 

And don't forget

do { v.resize(42); } while (!(i=42)); 

And the favorite

(i=42) ? v.resize(42) : i;

Or (the only serious c++ in the post)

int i(0);
std::vector<int> v(i=42);

Come on, this has no end

.....