c++ array assignment of multiple values
C++ArraysProgramming LanguagesSyntaxC++ Problem Overview
so when you initialize an array, you can assign multiple values to it in one spot:
int array [] = {1,3,34,5,6}
but what if the array is already initialized and I want to completely replace the values of the elements in that array in one line
so
int array [] = {1,3,34,5,6}
array [] = {34,2,4,5,6}
doesn't seem to work...
is there a way to do so?
C++ Solutions
Solution 1 - C++
There is a difference between initialization and assignment. What you want to do is not initialization, but assignment. But such assignment to array is not possible in C++.
Here is what you can do:
#include <algorithm>
int array [] = {1,3,34,5,6};
int newarr [] = {34,2,4,5,6};
std::copy(newarr, newarr + 5, array);
However, in C++0x, you can do this:
std::vector<int> array = {1,3,34,5,6};
array = {34,2,4,5,6};
Of course, if you choose to use std::vector
instead of raw array.
Solution 2 - C++
You have to replace the values one by one such as in a for-loop or copying another array over another such as using memcpy(..)
or std::copy
e.g.
for (int i = 0; i < arrayLength; i++) {
array[i] = newValue[i];
}
Take care to ensure proper bounds-checking and any other checking that needs to occur to prevent an out of bounds problem.
Solution 3 - C++
I made a little template function to conveniently assign values to a raw pointer.
template <typename T, typename U>
void set_array(T* array, U x) {
*array = x;
}
template <typename T, typename U, typename... V>
void set_array(T* array, U x, V... y) {
*array = x;
set_array(array + 1, y...);
}
An example is:
int main() {
int64_t array[10] = {};
set_array(array, 11, 12, 13, 14, 15, 16);
for (int i = 0; i < sizeof(array) / sizeof(array[0]); ++i) {
std::cout << array[i] << ", ";
}
}
...and it should print:
11, 12, 13, 14, 15, 16, 0, 0, 0, 0,
Solution 4 - C++
const static int newvals[] = {34,2,4,5,6};
std::copy(newvals, newvals+sizeof(newvals)/sizeof(newvals[0]), array);