c++ - terminate called after throwing an instance of 'std::out_of_range' -

why happen program says has no errors when run terminate called after throwing instance of 'std::out_of_range' what(): vector:_m_range_check. new c++ don't understand these errors

#include <vector> #include <iostream> #include <random> #include <time.h>  using namespace std; using std::vector;  int main() {  vector<int> deck; vector<int> nums; default_random_engine eng(time(0)); uniform_int_distribution<int> dis(0, 51);  int pos1; int pos2; int num1; int num2; int i; int n; int m;  (i = 0; < 52; i++) {     nums.push_back(i);  }  for(int j = 0; j < 52; j++) {     cout << nums.at(i) << "\n"; }   for(n = 0; n < 50; n++) {     pos1 = dis(eng);     pos2 = dis(eng);      cout << pos1 << "\n" << pos2 << "\n";      num1 = deck.at(pos1);     num2 = deck.at(pos2);  }  } 

it looks me if due typo, , should use variable 'j' in second loop. after first loop,

for (i = 0; < 52; i++) {     nums.push_back(i); } 

the variable 'i' contains value 52, sounds expected calling nums.at(i) throw std::out_of_range, since nums contains 52 values, starting @ index 0.

for(int j = 0; j < 52; j++) {     cout << nums.at(i) << "\n"; } 

fix replacing argument of at() 'j', assume original intent:

for(int j = 0; j < 52; j++) {     cout << nums.at(j) << "\n"; }