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    <p>You can pass an array by value, but you have to first wrap it in a struct or class. Or simply use a type like std::vector.</p> <p>I think the decision was for the sake of efficiency. One wouldn't want to do this most of the time. It's the same reasoning as why there are no unsigned doubles. There is no associated CPU instruction, so you have to make what's not efficient very hard to do in a language like C++. </p> <p>As @litb mentioned: "C++1x and boost both have wrapped native arrays into structs providing std::array and boost::array which i would always prefer because it allows passing and returning of arrays within structs"</p> <p>An array is a pointer to the memory that holds that array and the size. Note it is not the exact same as a pointer to the first element of the array.</p> <p>Most people think that you have to pass an array as a pointer and specify the size as a separate parameter, but this is not needed. You can pass a reference to the actual array itself while maintaining it's sizeof() status.</p> <pre><code>//Here you need the size because you have reduced // your array to an int* pointing to the first element. void test1(int *x, int size) { assert(sizeof(x) == 4); } //This function can take in an array of size 10 void test2(int (&amp;x)[10]) { assert(sizeof(x) == 40); } //Same as test2 but by pointer void test3(int (*x)[10]) { assert(sizeof(*x) == 40); //Note to access elements you need to do: (*x)[i] } </code></pre> <p>Some people may say that the size of an array is not known. This is not true. </p> <pre><code>int x[10]; assert(sizeof(x) == 40); </code></pre> <p>But what about allocations on the heap? Allocations on the heap do not return an array. They return a pointer to the first element of an array. So new is not type safe. If you do indeed have an array variable, then you will know the size of what it holds. </p>
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