Question

void swap(char *a,char *b){
  char t;
  t = *a;
  *a = *b;
  *b = t;
}
int main(void){
  char  a = '1';
  char b = '2';

  swap(&a,&b);
  printf("The value is %c and %c respectively\n",a,b);
  return 0;
}

in the above code, there’s a spot that confuse me

I think if a is a pointer, and *a is the value it points to

int *ptr, a = 1;

ptr = &a;
printf("The value of *ptr should be a: %d\n",*ptr);
printf("The value of *a should be an hex address: %p\n",ptr);

so in the swap(char *a, char *b) function,it takes the value not pointer( *a not a),

swap(&a, &b)
but it actually pass the pointer value to it as the parameter, and the code works. Anybody can explain it to me?(I think for swap(char *a){...} part, the declaration doesn’t mean it require *a to pass in, it means declare a pointer value a, not the value a points to as *ain elsewhere means).

Answer 1

* is confusing because it means two different, but closely related, things. In a variable declaration, * means “pointer”. In an expression, * means “dereference the pointer”.

It’s intended to be a helpful mnemonic: if you have char *a in your code it means that *a is a char.