c++ - Passing pointer to dynamically allocated array by copy to function has unexpected result -

i messing around passing pointers functions wrap head around how works , came across behavior unexpected. have following code:

#include <iostream> #include <string> #include <fstream> #include <sstream> #include <cmath> #include <iomanip>  using namespace std;  struct t {     string x;     string y; };  void foo(t*);  int main() {      t* ts = new t[2];     ts[0].x = "t1.x";     ts[0].y = "t1.y";     ts[1].x = "t2.x";     ts[1].y = "t2.y";      foo(ts);       cout << ts[0].x << endl;  }  void foo(t* s) {     delete[] s;     s = new t[2];     s[0].x = "foo.x";     s[1].y = "foo.y"; } 

the output here, interestingly enough, "foo.x". expected since inside of foo, s copy of pointer ts when delete[] s delete[] ts both point same address. s = new t[2]; should have no effect on ts. after foo returns, should no longer have access s or array points , ts should point knows where. missing somehthing?

note: test project made write , erase blocks of code test different concepts. includes , using namespace std ease of use, , not code writing sort of practical use, purely educational. also, using ms vs 2013.

try changing foo() , see result:

void foo(t* s) {   delete[] s;    // additional memory allocation   t* u = new t[2];    s = new t[2];   s[0].x = "foo.x";   s[1].y = "foo.y"; } 

by adding memory allocation, moved s location in memory, not anymore overlapping ts. otherwise, s allocated @ same location ts resided.

as pointed out in comments, observing undefined behavior, should no means rely on. example above illustrates pretty well.