从C中的函数返回多个值

Question

int getPositionsH(int r, int ans, int size){
    int x=0;
    int y=0;
        if (ans==9){
            x =0;
        } else
            x=(rand()%size);

        y=(rand()%10);  
    return x;
    return y;
}

Basically this is a function in c that is supposed to return 2 randomly generated positions x and y. However while debugging I noticed that x and y are still empty after this is executed. No idea why because I wrote return and everything. Any ideas why? Any help is appreciated.

Answer 1

A function can return only one value in C. As it is, the function returns only x and the statement return y; has no effect -- it's unreachable code.

If you want to return multiple values, you can either pass pointers and return the values in their content or make a struct to and return values.

typedef struct position {
   int x;
   int y;
}pos_type;

pos_type getPositionsH(int r, int ans, int size){
    pos_type p;
    int x=0;
    int y=0;
        if (ans==9){
            x =0;
        } else
            x=(rand()%size);

        y=(rand()%10);  

    p.x = x;
    p.y = y;
    reutrn p;
 }

and in the caller:

pos_type t = getPositionsH(...);

int x = t.x;
int y = t.y;
.....

Answer 2

you can't return two values this way.

int getPositionsH(int r, int ans, int size)

being declared as int return value will return only a single int

 return x;
 return y;

After returning x, program execution will return from the function, thus return y will remain unreachable.

Answer 3

Use pointers for x and y . Instead of returning values, the function is setting the values:

void getPositionsH(int r, int ans, int size, int *x, int *y) {
    *x = ans == 9 ? 0 : rand() % size;
    *y = rand() % 10;
}

to be called

int x,y;
getPositionsH(r, ans, size, &x, &y);
// use x and y as you wish...
int total = x + 7 * y;

getPositionsH is given two pointers (the address) to x and y. Using *x = sets the value of x (declared before the function call).

---

Note

The

  *x = ans == 9 ? 0 : rand() % size; 

statement is equivalent to

  if (ans == 9) *x = 0; else *x = rand() % size; 

Answer 4

return x; return y;

you have used return x . This function will return value to the calling function Now it will not come back and again return execute return y .Whenever you use return statement, you use keep in mind what actually return statement do ?

The return statement terminates the execution of a function and returns control to the calling function

So possibly you could refer this SO Q&A for solution of your problem. Other answers are also quite nice.