4 ms·

Yes, because "a" is not an object, it is a reference to an object. This reference is passed by value. This kind of thing is why I'd recommand everybody to know

• teamonkey 9y ago
  Could you give an example in C of how the above works?
  • pjmlp 9y ago
    Something like this, without any kind of error checking being done. #include <stdlib.h> // assume some hash table library, exercise for the reader typedef struct{} *hashtable_t; extern hashtable_t hashtable_init(void); extern void hashtable_put(hashtable_t hashtable, const char* key, const char* value); extern void hashtable_dump(hashtable_t hashtable); typedef struct { hashtable_t table; } data_t; void foo(data_t* a) { hashtable_put(a->table, "otherKey", "b"); } int main(void) { // let a = {key: 'a'}; data_t* a = (data_t*) malloc(sizeof(data_t)); a->table = hashtable_init(); hashtable_put(a->table, "key", "a"); foo(a); // console.log( a ); hashtable_dump(a->table); return 0; } The function foo() gets the numeric value of the a pointer, thus a parameter inside foo() points to the same memory location. If C had pass-by-reference, it would be possible to give (implicitly) the memory location of the local variable a in main() instead. For example, like in Pascal (var) or C++ (& in function declaration).