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C++: Pass array created in the function call line

开发者 https://www.devze.com 2023-02-11 05:05 出处:网络
How can I achieve a result like somebody would expect it according to the following code example: // assuming: void myFunction( int* arr );

How can I achieve a result like somebody would expect it according to the following code example:

// assuming: void myFunction( int* arr );

myFunction( 开发者_如何学编程[ 123, 456, 789 ] );

// as syntactical sugar for...

int values[] = { 123, 456, 789 };
myFunction( values );

The syntax I thought would work spit out a compile error.

  • How can I define an argument array directly in the line where the function is called?


If you were using C (C99, specifically), you could have used a compound literal here:

myFunction( (int []) {123, 456, 789} );

But the are not part of C++ (your compiler may still support them, though). In C++0x, you could use an initializer list, but they don't decay to pointers (not that using raw pointers is a good idea to begin with).

Although if you change the signature of myFunction to void myFunction( const int* arr ), you would be able to do the contrived call

myFunction( std::initializer_list<int>({123, 456, 789}).begin() );


Something like what you are looking for is proposed as part of the C++0x standard. I don't believe there is a way to do it otherwise (currently).

void function_name(std::initializer_list<float> list);

function_name({1.0f, -3.45f, -0.4f});


You can't achieve that syntactic sugar with C++ code, not in C++03. The only place where braces are allowed, referring to elements of an array, is in initializing a named array:

int arr[] = { 0, 1, 2 };

Nowhere else. You can however achieve a similar effect with Boost.Assign:

void f(const std::vector<int>& v)
{
}

// ...
f(boost::assign::list_of(123)(456)(789));


You can do that, but you have to use two things:

  1. use std::vector (or any other C++ STL class)

  2. use -std=c++0x compilation switch (or anything else for you compiler to use the c++0x standard).

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