为什么这个C程序的输出是这样的？

Question

Why the output of the following program is

x = 10 y = 18

?

int y;
void fun(int x) {
 x+=2;
 y=x+2;
}

int main() {
 int x;
 x=10; y=11;
 fun(x);
 fun(y);
 printf("x=%d y=%d\n", x,y);
 return 0;
}

Shouldn't the output be 10 and 11 ?

Answer 1

Since y is a global variable , in the first call fun(x); y becomes 14 since x is 10 , x += 2 makes x == 12 and then y = x + 2 which gives 14 . Then you call it with y == 14 , which makes the local x in fun() , x == 16 and then y == y + 2 which is 18 .

Answer 2

These are the states of variables before and after each of those function calls.

PRE: x=10, y=11
 fun(x);
POST: x=10, y=14
PRE: x=10, y=14
 fun(y);
POST: x=10, y=18

---

If you simply rename the local variable within fun() to something other than x, it becomes less complicated.

void fun(int x) {
  x+=2;
  y=x+2;
}

can be rewritten as:

void fun(int local_var) {
  y=local_var+4; //y is global, local_var is thrown away at the end of this scope.
}