如何修复 C++ 中的分段错误（核心转储）？

Question

I'm writing a program that combines 2 vectors and sorts them and then prints the vector but I'm not using a third vector. Instead I'm combining one vector with another and then sorting the combined vector. But I get a error called "Segmentation fault".

here's the code:

#include<bits/stdc++.h>
using namespace std;

int main() {
    // your code goes here
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    int m,n; 
    cin >> m >> n;
    vector<int> nums1, nums2;
    for(int i=0; i<m; i++) cin >> nums1[i];
    for(int i=0; i<n; i++) cin >> nums2[i];
    for(int i=0; i<n; i++){ // nums2 
        nums1.push_back(nums2[i]); // I am adding all the elements present in nums2 into nums1
    }
    sort(nums1.begin(), nums1.end());
    for(int i=0; i<(m+n); i++) cout << nums1[i] << " ";
    return 0;
}

The error I get: run: line 1: 3 Segmentation fault (core dumped) LD_LIBRARY_PATH=/usr/local/gcc-8.3.0/lib64./a.out

Please tell me how I can fix this error and how I could avoid it in the future.

Answer 1

Here:

vector<int> nums1, nums2;

for(int i=0; i<m; i++) cin >> nums1[i]; // this causes undefined behavior
for(int i=0; i<n; i++) cin >> nums2[i]; // also this one

your vectors have no buffer to store data so you need to do this before using operator[] :

vector<int> nums1(m), nums2(n);

nums1.push_back(2); // will add 2 to the back of nums1 so size will become m + 1
nums2.push_back(6); // will add 6 to the back of nums2 so size will become n + 1

// you can do as many push_backs as you want until
// your computer runs out of memory

Now both will be initialized with m and n number of elements respectively.

If you used the at function instead of [] , the program would throw a std::out_of_range exception and you would suddenly notice that you were trying to go out of bounds. But of course at comes at a performance cost.