C++ – Pro/con: Initializing a variable in a conditional statement

c++coding-styleif-statement

In C++ you can initialize a variable in an if statement, like so:

if (CThing* pThing = GetThing())
{
}

Why would one consider this bad or good style? What are the benefits and disadvantages?

Personally i like this style because it limits the scope of the pThing variable, so it can never be used accidentally when it is NULL. However, i don't like that you can't do this:

if (CThing* pThing = GetThing() && pThing->IsReallySomeThing())
{
}

If there's a way to make the above work, please post. But if that's just not possible, i'd still like to know why.

Question borrowed from here, similar topic but PHP.

Best Solution

The important thing is that a declaration in C++ is not an expression.

bool a = (CThing* pThing = GetThing()); // not legit!!

You can't do both a declaration and boolean logic in an if statement, C++ language spec specifically allows either an expression or a declaration.

if(A *a = new A)
{
    // this is legit and a is scoped here
}

How can we know whether a is defined between one term and another in an expression?

if((A *a = new A) && a->test())
{
    // was a really declared before a->test?
}

Bite the bullet and use an internal if. The scope rules are useful and your logic is explicit:

if (CThing* pThing = GetThing())
{
    if(pThing->IsReallySomeThing())
    {
    }
}

Related Solutions

Should a function have only one return statement

I often have several statements at the start of a method to return for "easy" situations. For example, this:

public void DoStuff(Foo foo)
{
    if (foo != null)
    {
        ...
    }
}

... can be made more readable (IMHO) like this:

public void DoStuff(Foo foo)
{
    if (foo == null) return;

    ...
}

So yes, I think it's fine to have multiple "exit points" from a function/method.

C++ – What are the differences between a pointer variable and a reference variable in C++

A pointer can be re-assigned:

int x = 5;
int y = 6;
int *p;
p = &x;
p = &y;
*p = 10;
assert(x == 5);
assert(y == 10);

A reference cannot be re-bound, and must be bound at initialization:

int x = 5;
int y = 6;
int &q; // error
int &r = x;

A pointer variable has its own identity: a distinct, visible memory address that can be taken with the unary & operator and a certain amount of space that can be measured with the sizeof operator. Using those operators on a reference returns a value corresponding to whatever the reference is bound to; the reference’s own address and size are invisible. Since the reference assumes the identity of the original variable in this way, it is convenient to think of a reference as another name for the same variable.
```
int x = 0;
int &r = x;
int *p = &x;
int *p2 = &r;

assert(p == p2); // &x == &r
assert(&p != &p2);
```

You can have arbitrarily nested pointers to pointers offering extra levels of indirection. References only offer one level of indirection.

int x = 0;
int y = 0;
int *p = &x;
int *q = &y;
int **pp = &p;

**pp = 2;
pp = &q; // *pp is now q
**pp = 4;

assert(y == 4);
assert(x == 2);

A pointer can be assigned nullptr, whereas a reference must be bound to an existing object. If you try hard enough, you can bind a reference to nullptr, but this is undefined and will not behave consistently.

/* the code below is undefined; your compiler may optimise it
 * differently, emit warnings, or outright refuse to compile it */

int &r = *static_cast<int *>(nullptr);

// prints "null" under GCC 10
std::cout
    << (&r != nullptr
        ? "not null" : "null")
    << std::endl;

bool f(int &r) { return &r != nullptr; }

// prints "not null" under GCC 10
std::cout
    << (f(*static_cast<int *>(nullptr))
        ? "not null" : "null")
    << std::endl;

You can, however, have a reference to a pointer whose value is nullptr.

Pointers can iterate over an array; you can use ++ to go to the next item that a pointer is pointing to, and + 4 to go to the 5th element. This is no matter what size the object is that the pointer points to.
A pointer needs to be dereferenced with * to access the memory location it points to, whereas a reference can be used directly. A pointer to a class/struct uses -> to access its members whereas a reference uses a ..
References cannot be put into an array, whereas pointers can be (Mentioned by user @litb)
Const references can be bound to temporaries. Pointers cannot (not without some indirection):
```
const int &x = int(12); // legal C++
int *y = &int(12); // illegal to take the address of a temporary.
```
This makes const & more convenient to use in argument lists and so forth.

Best Solution

Related Solutions

Should a function have only one return statement

C++ – What are the differences between a pointer variable and a reference variable in C++

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