References and function calls

Hi

We know that the variables when making a function call are copied using assignment, i.e. in

Code:

---

void test(int a)
{
cout << a;
}

int main()
{
test(3)
return 0;
}

---

we have

Code:

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int a;
a=3;

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But what if I had chosen to call test using a reference? How would the reference be passed to the function test? I'm thinking of the fact that references need to be initialized.

Best,
Niles.

If you tried it with a non-const reference parameter, you should get a compile error. If you tried it with a const reference parameter, it should work just fine.

Quote:

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Originally Posted by laserlight

If you tried it with a non-const reference parameter, you should get a compile error. If you tried it with a const reference parameter, it should work just fine.

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But how is it passed? If I had

Code:

---

void test(int &a)
{
cout << a;
}

int main()
{
int b=3;
test(b)
return 0;
}

---

,

then it does not get passed as

Code:

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int a&; //ERROR
a=b; //ERROR

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Then how does it get passed by assignment?

With a const reference parameter, the reference is initialised to refer to the argument even if the argument is a literal. It is quite safe since due to the constness you would not be modifying the literal through the reference.

I'm gonna guess:

Code:

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int & a = b;

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Not 100% sure what you're asking here about "how it gets passed". It's almost like you're asking how the assignment/initialization of the reference parameter happens.

Quote:

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Originally Posted by laserlight

If you tried it with a non-const reference parameter, you should get a compile error. If you tried it with a const reference parameter, it should work just fine.

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No it shouldn't!!

Quote:

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Originally Posted by hk_mp5kpdw

I'm gonna guess:

Code:

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int & a = b;

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Not 100% sure what you're asking here about "how it gets passed". It's almost like you're asking how the assignment/initialization of the reference parameter happens.

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That is an initialization; arguments in C++ are passed to function parameters by assignment. That is my question: Since references cannot be assigned, but must be initialized, then how can I have a reference as a function argument?

I am not sure I understand Laserlight's response in light of my question. But you say it gets initialized?

Quote:

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Originally Posted by Sipher

No it shouldn't!!

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Notice that I was talking with reference to Niels_M's code example in post #1. The code example in post #3 is fine in that respect (though it contains a syntax error), since the variable b rather than the integer literal 3 was passed to test.

Quote:

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Originally Posted by Niels_M

arguments in C++ are passed to function parameters by assignment.

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That is false.

O_O ... Sorry, my bad. I didn't notice it.

Quote:

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Originally Posted by laserlight

That is false.

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?!

I've read numerous places that it is the case, the C-FAQ being one of them (as far as I remember).

Quote:

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Originally Posted by Niels_M

I've read numerous places that it is the case, the C-FAQ being one of them (as far as I remember).

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Then all those places are wrong, or they are using "assignment" in its more general computer science meaning.

Quote:

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Originally Posted by C++03 Section 5.2.2 Paragraph 4 (part)

When a function is called, each parameter shall be initialized with its corresponding argument.

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Quote:

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Originally Posted by laserlight

Then all those places are wrong, or they are using "assignment" in its more general computer science meaning.

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Ok, thanks for that. Just as a reference: 10.6 Arrays and Pointers as Function Arguments