C 中的赋值混淆

Question

I am new in C so I need some help.

for example I have this struct object

struct Frac {
   int num; 
   int dec
}

which i do this in c:

struct Frac fract1 = {1, 2};
struct Frac temp; 
temp = fract1;

Is temp point to a copy of fract1 or is it pointing to the actual fract1? I tried it on computer by printing the address and it is pointing to a copy of fract1. If this is in Java, it will be pointing to the actual fract1. Do correct me if I am wrong. Can I ask why temp is pointing to a copy of fract1 in C instead of the actual object itself thanks!

Answer 1

Nothing is pointing to anything. Assignment copies values. Assignment always copies values. Sometimes those values are pointers, but if you don't see a pointer, there isn't a pointer. C doesn't have Java's dichotomy between simple types with value semantics and everything-else with reference semantics.

Answer 2

" Is temp point to a copy of fract1 or is it pointing to the actual fract1 "?

First of all, temp and fract1 are not pointers. There are object types of the structures.

The content of frac1 is copied to temp but temp is a different object.

Answer 3

Is temp point to a copy of fract1 or is it pointing to the actual fract1? I tried it on computer by printing the address and it is pointing to a copy of fract1.

You're correct. It would point to a copy of fract1.

If this is in Java, it will be pointing to the actual fract1.

You're correct, in Java there would be two references to the same object.

Can I ask why temp is pointing to a copy of fract1 in C instead of the actual object itself

C doesn't automatically allocate objects on the heap unless you ask it to. You could reproduce something similar to Java by using pointers and heap allocations. But note that you are required to free objects explicitly unlike in Java.

struct Fraction *frac1 = malloc(sizeof(struct Fraction));
frac1->num = 1;
frac1->dec = 2;

struct Fraction *temp; 
temp = frac1;

// Both frac1 and temp point to the same object
...

// Make sure you free the object eventually
free(frac1);

Answer 4

Yes, the assignment operator always creates a copy. This is what you would normally expect. The Java behaviour is actually the deviant one.

For example in C or Java,

int a = 3;
int b = a;

Would you expect b to point to the 'actual' a or would b be a copy of a containing the same value 3 ? This is exactly what C does, even for structures, unlike Java. For all non-trivial objects, Java implicitly allocates memory on the heap and creates a reference (equivalent to a pointer). What you call 'object' in Java is actually a reference to an object and when you copy it using the assignment operator, you are actually copying this reference. Allocation of composite objects on the stack that is possible in C is not possible in a language like Java.

If you want to create a reference, you can still do that using pointers in C. You will have to 'dereference' this pointer to actually retrieve the object behind it.

Modifying your original example:

struct Frac fract1 = {1, 2};
struct Frac *temp;
struct Frac *temp2; 
temp = &fract1; // Now temp points to the original fract1
temp2 = temp;   // Copying a reference: temp2 is a copy of temp but both point to the same fract1

temp->num = 3;  // This modifies fract1 indirectly

Answer 5

Both frac1 and temp are variables. These variables occupy two different spaces in memory.

What is going on in C, is that temp = frac1 is copying the contents of the memory named frac1 into the memory named temp .

If you are dealing with a pointer, these ideas of spaces in memory and copying the contents still applies.

For example:

int *p; - there is a space in memory named p; however, the contents of this space in memory contain an address. This address in turn (likely) refers to a different space in memory. Thus p = ... will update the contents of p (which is a pointer) so that it will point to someplace new. Meanwhile *p = ... changes the contents of the memory to which the contents of p pointed.