I would like to know - is there any way to copy array to a function in C/C++? Not pass it just an address, but create a copy... Like for int value. For example here we send not a copy(((:
void arrayAbs(int array[], int size) {
for ( int i = 0; i < size; i++ ) {
if ( array[i] < 0 ) {
array[i] *= -1;
}
}
}
This is not possible for arrays - at least when the size of the array is not known at compile time. Since you are passing size as a separate parameter, it appears that a solution with a fixed-size array would not work in your situation.
However, you can use std::vector<int> instead. This container will be copied when you pass it by value. As an additional benefit, it knows its size, so passing int size would become unnecessary.
C and C++ arrays are not copyable. But you can use an array-like standard library container type, such as std::array:
template <size_t N>
void arrayAbs(std::array<int, N> a)
{
for ( int i = 0; i < a.size(); i++ ) {
if ( a[i] < 0 ) {
a[i] *= -1;
}
}
}
Here, a is passed by value, so it is a copy of the array passed in by the caller.
You can not do this with arrays. Either arrays are converted to pointers to their first elements when passed to a function (in C and C++) or if the parameter is defined as a reference to an array then the reference is passed. in any case neither copy of an array is created.
In C++ you could use standard class std::array. in this case you indeed could work with a copy of the original array. In C you could use an analogy of std::array in C++ by wrapping an array in a structure.
size as a separate parameter. The implication would be that his arrayAbs would become generic on the size of the std::array passed into it (see Juan's answer).In C++: Use std::array. In C: Wrap it in a struct.
typedef struct {
int array[10];
} wrapper;
wrapper function(wrapper wrapped_array) {
wrapped_array.array[0] = 42;
return wrapped_array;
}
int main() {
wrapper wa;
wrapper wa2 = function(wa);
}
std::array<T>(see reference), orstd::vector<T>std::vectoror (better)std::array(C++11 only)