# a = b = new c?

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Waaayoff    952

If a and b are pointers, does this

A) Allocate memory once for b and also store address in a

B) Allocate memory twice, once for a and once for b

Waaayoff    952

Ok thanks :)

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NightCreature83    5002

You should never write code like that as this doesn't mean the same, and the way that statement is written will often degenerate into this:

int* a = b = new c();


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Bregma    9202

The assignment operator binds left, which means

a = b = new c;

is the same as

a = (b = new c);


or

a.operator=(b.operator=(c.operator new()));


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frob    44916

It is also the same as:

b = new c;

a = b;

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Vortez    2714

What would be the point of such code??? One pointer is enough imo...

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frob    44916

What would be the point of such code??? One pointer is enough imo...

Depends on the algorithm.

There are many that require two pointers, one that will always be there and another that gets modified.  In this case they are initializing both pointers at once.

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Waaayoff    952

What would be the point of such code??? One pointer is enough imo...

Depends on the algorithm.

There are many that require two pointers, one that will always be there and another that gets modified.  In this case they are initializing both pointers at once.

Yep that's why i'm doing it

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Camilo    189

[...]

Yep that's why i'm doing it

As a pedagogical exercise it's an interesting question, but don't use it. If you have to ask on a forum what it does the next person that sees your code (or yourself in a few months) might not know, either. Not worth it, just to save 1 line of code.

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Tom KQT    1704

As a pedagogical exercise it's an interesting question, but don't use it. If you have to ask on a forum what it does the next person that sees your code (or yourself in a few months) might not know, either. Not worth it, just to save 1 line of code.

Exactly. This will not make the code run faster. It even won't make the code more readable (quite the opposite), so why bother?

I personaly would definitely put the a = b part on a separate row and maybe also add a short comment why am I requesting another copy of the pointer (or use such a name that would suggest it clearly).

Edited by Tom KQT

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Waaayoff    952

What about when you're doing something like this:

ShaderParameter* ParameterManager::CreateFloat(const std::string& Name)
{

if (!pParam)
{
mParameters[Name] = pParam = new FloatParameter;
}

return pParam;
}
Edited by Waaayoff

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Brother Bob    10344

What about when you're doing something like this:

ShaderParameter* ParameterManager::CreateFloat(const std::string& Name)
{

if (!pParam)
{
mParameters[Name] = pParam = new FloatParameter;
}

return pParam;
}

What are you gaining over just having two separate assignment expressions?

pParam = new FloatParameter;
mParameters[Name] = pParam;


Trying to be clever when it comes to coding is rarely a benefit over being clear. The difference is purely in the information you convey to the programmer; the compiler couldn't care less. Your code does not have to be clear because it benefits the compiler, it has to be clear because YOU, the programmer, has to read the code.

Edited by Brother Bob

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Waaayoff    952

Well it's not an attempt at optimization i assure you. Also, I didn't realize it was that unclear to read...

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Tom KQT    1704

Also, I didn't realize it was that unclear to read...

Then why did you create this thread? ;-)

You weren't sure how it works, that's IMHO a good proof that it isn't clear to read (at least for you and as it's your code...).

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Waaayoff    952

Also, I didn't realize it was that unclear to read...

Then why did you create this thread? ;-)

You weren't sure how it works, that's IMHO a good proof that it isn't clear to read (at least for you and as it's your code...).

Yes but that's like saying i shouldn't use smart pointers because they're not readable for me since i don't know how they work.. I get what you guys are saying about readability but i just thought this was a simple assignment operation that was common knowledge for non beginners (unlike me)

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m3mb3rsh1p    440

There's also the type casting taking place:

T a = b = (T)new c();
mParameters[Name] = pParam = (ShaderParameter)new FloatParameter;


...or however it should be done.

I think it would be clearer to have the cast or temporary identified on a separate line.

Edited by m3mb3rsh1p

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Serapth    6671

a = b = new c creates (allocates) one object of type c, then assigns the pointer to b, and then assigns the same pointer to a.

Careful, that isn't actually true.

a will point to b, while b will point to c.  The fact that b currently points to c wont hold true if b is reassigned.

... I think I said that poorly.

EDIT: ------------------------------------------------------ IGNORE EVERYTHING I SAY IN THIS THREAD!!!! -----------------------------------------------------------------

EDIT: Seriously... do it.  Took a double dose of stupid pills this afternoon.  I'd edit out my comments for the sake of confusion, but thats makes things worse.

Edited by Serapth

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Khatharr    8812

a = b = new c creates (allocates) one object of type c, then assigns the pointer to b, and then assigns the same pointer to a.

Careful, that isn't actually true.

a will point to b, while b will point to c.  The fact that b currently points to c wont hold true if b is reassigned.

... I think I said that poorly.

http://objection.mrdictionary.net/go.php?n=6774974

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Bregma    9202

There's also the type casting taking place:

T a = b = (T)new c();
mParameters[Name] = pParam = (ShaderParameter)new FloatParameter;


...or however it should be done.

I think it would be clearer to have the cast or temporary identified on a separate line.

Er, no.  There is no casting.

Assuming a and b are of type pointer to c, then assigning the value of b to a requires no type cast, and the expression new c returns a value of type pointer to c, so again there is no type cast.

Again, assuming pParam is of type ShaderParameter* and mParameters is of type std::map<std::string, ShaderParameter*> and that FloatParameter is a class derived from ShaderParameter, there is no implicit type cast anywhere in the expression.  Of course, those assumptions may not hold true, in which case the code woul be unlikely to compile anyway because it's missing explicit casts.

So, I think adding explicit casts where no cast is required would confuse the reader.  I don't have a problem with the multiple expressions on the same line, but then I've had to maintain maybe a million lines of C code over the years and recognize the idiom.  There are many things that are far more important to readability than that (eg. max 80 character lines).

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Serapth    6671

a = b = new c creates (allocates) one object of type c, then assigns the pointer to b, and then assigns the same pointer to a.

Careful, that isn't actually true.

a will point to b, while b will point to c.  The fact that b currently points to c wont hold true if b is reassigned.

... I think I said that poorly.

http://objection.mrdictionary.net/go.php?n=6774974

Actually that captures it perfectly ( and is completely wrong! ;) ).

The address of b is assigned to a, not the value.  This is exactly what I was warning of.

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Brother Bob    10344

a = b = new c creates (allocates) one object of type c, then assigns the pointer to b, and then assigns the same pointer to a.

Careful, that isn't actually true.

a will point to b, while b will point to c.  The fact that b currently points to c wont hold true if b is reassigned.

... I think I said that poorly.

http://objection.mrdictionary.net/go.php?n=6774974

Actually that captures it perfectly ( and is completely wrong! ;) ).

The address of b is assigned to a, not the value.  This is exactly what I was warning of.

int main()
{
int *a, *b;

a = b = new int;

std::cout << "value of a is " << a << std::endl;
std::cout << "value of b is " << b << std::endl;
std::cout << "address of b is " << &b << std::endl;
}


Is the value of a equal to the value of b or to the address of b?

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Serapth    6671

a = b = new c creates (allocates) one object of type c, then assigns the pointer to b, and then assigns the same pointer to a.

Careful, that isn't actually true.

a will point to b, while b will point to c.  The fact that b currently points to c wont hold true if b is reassigned.

... I think I said that poorly.

http://objection.mrdictionary.net/go.php?n=6774974

Actually that captures it perfectly ( and is completely wrong! ;) ).

The address of b is assigned to a, not the value.  This is exactly what I was warning of.

int main()
{
int *a, *b;

a = b = new int;

std::cout << "value of a is " << a << std::endl;
std::cout << "value of b is " << b << std::endl;
std::cout << "address of b is " << &b << std::endl;
}


Is the value of a equal to the value of b or to the address of b?

The value of a and b are the same ( for now ), the address of a and b are different.

The address pointed to by a and b will be the same (for now), the address OF a and b will be different.

Edited by Serapth

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Brother Bob    10344

So the conclusion is that the value of b is assigned to a?