c++ array assignment of multiple values

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?

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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.

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最佳答案

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.

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const static int newvals[] = {34,2,4,5,6};


std::copy(newvals, newvals+sizeof(newvals)/sizeof(newvals[0]), array);
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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,

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