Difference between C and Java when it comes to variables?

Question

I recently asked this question: Why aren't my pointers to integer arrays working?

One of the suggestions was to use regular variables instead of pointers pointing to variables.

I've programmed in Java in the past. If I understand correctly, if you have this code in Java:

void test(b) {
    b++;
}

void main() {
    int a = 0;
    test(a);
}

Then a would remain at 0. But if I understood correctly, in C, if I have this code:

void test(b) {
    b++;
}

int main() {
    int a = 0;
    test(a);
    return 0;
}

Then a becomes 1?

Neither of these is valid code, so it's somewhat tricky to talk about how they would behave... — Oliver Charlesworth, Apr 01 '13 at 17:56
it would only work that way if you sent a pointer to a, not a itself. In the code you have, you're making a copy of a that's in the function's scope, but it disappears when the function ends. — Ken, Apr 01 '13 at 17:57
@OliCharlesworth It's valid on older C versions because of default `int`. — Mysticial, Apr 01 '13 at 17:58

score 4 · Accepted Answer · answered Apr 01 '13 at 17:57

4

No , in c passing a value makes a copy of that value, and the original variable isn't changed.

Now, you can pass a value of the address of that variable, then in the function you look where the passed address is pointing at, and change the value at that address and voila the original variable is changed.

answered Apr 01 '13 at 17:57

So how would you explain doing this: http://stackoverflow.com/a/15748364/1251385 without pointers? – user1251385 Apr 01 '13 at 17:59
2

@user: There are no pointers there. Arrays become pointers when passed to functions. – Oliver Charlesworth Apr 01 '13 at 18:01
@user1251385 In short, you can't in C. In C++ you can use references. In Java, I don't think it's possible at all on primitive types. – Mysticial Apr 01 '13 at 18:01
1

"Arrays become pointers when passed to functions." Wow. Really? Thanks C devs for being so confusing! – user1251385 Apr 01 '13 at 18:02
@Armin an array is not an address... – Oliver Charlesworth Apr 01 '13 at 18:04
Anyone looking for more info: http://stackoverflow.com/a/1106977/1251385 – user1251385 Apr 01 '13 at 18:14
@user1251385: Check [this document](http://www.open-std.org/jtc1/sc22/wg14/www/docs/n1570.pdf), § 6.3.2.1 ¶ 3, and [this document](http://cm.bell-labs.com/who/dmr/chist.html) in the section titled "Embryonic C". – John Bode Apr 01 '13 at 18:39

score 2 · Answer 2 · answered Apr 01 '13 at 18:00

2

No. In the C example, the a would still retain its original value. In C, all parameters are passed by value, while in Java, only primitive types are passed by value, all others are passed by reference. To achieve the effect you describe, your code should look like this:

void test(int* b) {
    (*b)++;
}

int main() {
   int a = 0;
   test(&a);
   return 0;
}

Here, test accepts a pointer to int which it then dereferences (gets the value that lies in the memory on the specified address) and increases it by 1. In main() you pass a address of a.

answered Apr 01 '13 at 18:00

Andrej Palicka

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In Java, "all others" are passed by value as well, just value of the reference. – jedwards Apr 01 '13 at 18:03
2

Java is just like C in that there is no pass-by-reference: all types, both primitives and references/pointers, are passed *by value*. The difference in terminology (pass-by-reference vs pass-by-value) is important because there are languages that do have pass-by-reference. – Joni Apr 01 '13 at 18:07
Hmmm, I've been living in a lie my whole life :-) I thought that what Java does with its objects is in fact pass by reference. What does pass-by-reference means then? – Andrej Palicka Apr 01 '13 at 18:14
@akcilap You can think of it as pass-by-value, but Java only ever gives you pointers to objects. – Nicholas Wilson Apr 18 '13 at 09:52

score 1 · Answer 3 · answered Apr 01 '13 at 18:03

Wait a moment."In C, arguments are passed to functions by value while other languages may pass variables by reference. This means that the receiving function gets copies of the values and has no direct way of altering the original variables. For a function to alter a variable passed from another function, the caller must pass its address (a pointer to it), which can then be dereferenced in the receiving function. See Pointers for more information."

score 0 · Answer 4 · answered Apr 01 '13 at 18:03

In both languages, the passing of the variable is by copying in the value. It's just that with C, you can pass a memory location (pointer) as the value, and effectively get pass by reference.

In Java, you could do the same thing (without actually having a memory address value) by passing in a reference; however, java built-in types cannot be passed by reference as only objects can be referred to by reference. Note that arrays are actually objects in java, so some of the rules must be interpreted as objects when dealing with arrays, even if they are arrays of Java primitives.

Difference between C and Java when it comes to variables?

4 Answers4