在 C function 中将结构作为参数传递

Question

typedef struct { int b, p; } A;
void f(A c);
int main()
{
  int i;
  A a = {1,2};
  f(a);
  printf("%d,%d\n",a.b,a.p);
  return 0;
}
void f(A c)
{
  int j;
  c.b += 1;
  c.p += 2;
}

In the following code, why is it that ab and ap are still 1 and 2 even after being passed through f(a), which should change them to 2 and 3?

Answer 1

What you are using is called call by value, in call by value the changes you are doing to function arguments will not be reflected in the caller, for that you need call by reference

typedef struct { int b, p; } A;

void f(A c);

void f2(A *c);

int main()
{
  int i;
  A a = {1,2};

  // call by value
  f(a);
  printf("%d,%d\n",a.b,a.p);

  //call by referecne    
  f2(&a);

  printf("%d,%d\n",a.b,a.p);
  return 0;
}

void f(A c)
{
  int j;
  c.b += 1;
  c.p += 2;
}

void f2(A *c)
{
  int j;
  c->b += 1;
  c->p += 2;
}