Question

What is the difference between

(type)value

and

type(value)

in C++?

Was it helpful?

Solution

There is no difference; per the standard (§5.2.3):

A simple-type-specifier (7.1.5) followed by a parenthesized expression-list constructs a value of the specified type given the expression list. If the expression list is a single expression, the type conversion expression is equivalent (in definedness, and if defined in meaning) to the corresponding cast expression (5.4).

Since the question specified the difference between type(value) and (type)value, there is absolutely no difference.

If and only if you're dealing with a comma-separated list of values can there be a difference. In this case:

If the expression list specifies more than a single value, the type shall be a class with a suitably declared constructor (8.5, 12.1), and the expression T(x1, x2, ...) is equivalent in effect to the declaration T t(x1, x2, ...); for some invented temporary variable t, with the result being the value of t as an rvalue.

As Troubadour pointed out, there are a certain names of types for which the type(value) version simply won't compile. For example:

char *a = (char *)string;

will compile, but:

char *a = char *(string);

will not. The same type with a different name (e.g., created with a typedef) can work though:

typedef char *char_ptr;

char *a = char_ptr(string);

OTHER TIPS

There is no difference; the C++ standard (1998 and 2003 editions) is clear about this point. Try the following program, make sure you use a compiler that's compliant, such as the free preview at http://comeaucomputing.com/tryitout/.

#include <cstdlib>
#include <string>
int main() {
  int('A'); (int) 'A'; // obvious
  (std::string) "abc"; // not so obvious
  unsigned(a_var) = 3; // see note below
  (long const&) a_var; // const or refs, which T(v) can't do
  return EXIT_SUCCESS;
}

Note: unsigned(a_var) is different, but does show one way those exact tokens can mean something else. It is declaring a variable named a_var of type unsigned, and isn't a cast at all. (If you're familiar with pointers to functions or arrays, consider how you have to use a parens around p in a type like void (*pf)() or int (*pa)[42].)

(Warnings are produced since these statements don't use the value and in a real program that'd almost certainly be an error, but everything still works. I just didn't have the heart to change it after making everything line up.)

There is no difference when both are casts, but sometimes 'type(value)' is not a cast.

Here's an example from standard draft N3242, section 8.2.1:

struct S 
{
    S(int);
};

void foo(double a) 
{
    S w( int(a) ); // function declaration
    S y( (int)a ); // object declaration
}

In this case 'int(a)' is not a cast because 'a' is not a value, it is a parameter name surrounded by redundant parentheses. The document states

The ambiguity arising from the similarity between a function-style cast and a declaration mentioned in 6.8 can also occur in the context of a declaration. In that context, the choice is between a function declaration with a redundant set of parentheses around a parameter name and an object declaration with a function-style cast as the initializer. Just as for the ambiguities mentioned in 6.8, the resolution is to consider any construct that could possibly be a declaration a declaration.

In c there is no type (value), while in c/c++ both type (value) and (type) value are allowed.

To illustrate your options in C++ (only one has a safety check)

#include<boost/numeric/conversion/cast.hpp> 

using std::cout;
using std::endl;
int main(){

    float smallf = 100.1;

    cout << (int)smallf << endl; // outputs 100 // c cast
    cout << int(smallf) << endl; // outputs 100 // c++ constructor = c cast

    cout << static_cast<int>(smallf) << endl; // outputs 100
//  cout << static_cast<int&>(smallf) << endl; // not allowed
    cout << reinterpret_cast<int&>(smallf) << endl; // outputs 1120416563
    cout << boost::numeric_cast<int>(smallf) << endl; // outputs 100

    float bigf = 1.23e12;

    cout << (int)bigf << endl; // outputs -2147483648
    cout << int(bigf) << endl; // outputs -2147483648

    cout << static_cast<int>(bigf) << endl; // outputs -2147483648
//  cout << static_cast<int&>(bigf) << endl; // not allowed
    cout << reinterpret_cast<int&>(bigf) << endl; // outputs 1401893083
    cout << boost::numeric_cast<int>(bigf) << endl; // throws bad numeric conversion
}
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