C++ -- return x,y; What is the point? C++ -- return x,y; What is the point? c c

C++ -- return x,y; What is the point?


According to the C FAQ:

Precisely stated, the meaning of the comma operator in the general expression

e1 , e2

is "evaluate the subexpression e1, then evaluate e2; the value of the expression is the value of e2." Therefore, e1 had better involve an assignment or an increment ++ or decrement -- or function call or some other kind of side effect, because otherwise it would calculate a value which would be discarded.

So I agree with you, there is no point other than to illustrate that this is valid syntax, if that.

If you wanted to return both values in C or C++ you could create a struct containing x and y members, and return the struct instead:

struct point {int x; int y;};

You can then define a type and helper function to allow you to easily return both values within the struct:

typedef struct point Point;Point point(int xx, int yy){  Point p;  p.x = xx;  p.y = yy;  return p;}

And then change your original code to use the helper function:

Point foo(){  int x=0;  int y=20;  return point(x,y); // x and y are both returned}

And finally, you can try it out:

Point p = foo();printf("%d, %d\n", p.x, p.y);

This example compiles in both C and C++. Although, as Mark suggests below, in C++ you can define a constructor for the point structure which affords a more elegant solution.


On a side note, the ability to return multiple values directly is wonderful in languages such as Python that support it:

def foo():  x = 0  y = 20  return x,y # Returns a tuple containing both x and y>>> foo()(0, 20)


The comma in parameter lists is just there to separate the parameters, and is not the same as the comma operator. The comma operator, as in your example, evaluates both x and y, and then throws away x.

In this case, I would guess that it is a mistake by someone who tries to return two values, and didn't know how to do it.


The comma operator is primarily used in for statements like so:

for( int i=0, j=10; i<10; i++, j++ ){    a[i] = b[j];}

The first comma is not a comma operator, it's part of the declaration syntax. The second is a comma operator.