Why does the swap through pointer not work in vector in C++?

Question

I have the following code:

using namespace std; template<class T> void myswap(T* a, T* b) { T temp = *a; *a = *b; *b = temp; } template<class T> void printArr(vector<T> arr) { cout << "[ "; for(int i = 0; i < arr.size(); i++) { cout << arr[i] << " "; } cout << "]"; } void sampleSwap(vector<int> arr) { //swapping first two element myswap(&arr[0], &arr[1]); printArr(arr); cout << endl; } int main() { srand(500); vector<int> arr; for(int i = 0; i < N; i++) { arr.push_back(rand() % LIMIT); } printArr(arr); cout << endl; sampleSwap(arr); printArr(arr); return 0; }

I got the following output:

[ 398 57 417 270 271 918 903 314 569 70 ] [ 57 398 417 270 271 918 903 314 569 70 ] [ 398 57 417 270 271 918 903 314 569 70 ]

I tried to swap the first and second elements of my vector. The vector content is changed in the function but it doesn't retain the change after the function exits. Am I missing something?

You take a copy of you vector in sampleSwap, use pass by reference: sampleSwap(vector<int>& arr) — quantdev
– quantdev, Commented Feb 7, 2015 at 23:45

Baum mit Augen · Accepted Answer · 2015-02-07 23:46:29Z

You pass the vector by value to sampleSwap, so only a copy of the vector you have in the main gets modified. Pass by reference instead:

void sampleSwap(vector<int> &arr) ^

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