c ++中2d向量的cout缺少行

Question

I am quite insane, and cannot interpret what happens here.

#include <vector>
#include <algorithm>
#include <iostream>

using namespace std;

int main() {
    int i = 0, j = 0, m = 3, n = 3;
    vector<vector<int> > vvi(3, vector<int>(3, 1));

    // why the following code outputs only single row, 
    // i.e.,"111" from vvi[0]? what about vvi[1], vvi[2]?
    for(; i < m; ++i) {
        for(; j< n; ++j) {
            cout << vvi[i][j];
        }
    }
    // any difference from the code below? 
    // for(int i=0; i < m; ++i) {
    //   for(int j= 0; j< n; ++j) {
    //      cout << vvi[i][j];
    //   }
    // }
}

Answer 1

In the first iteration of the outer loop the counter variable j gets incremented from 0 to n , but since it never gets reset to 0 it stays at n , thus the inner loop condition j < n is false for all subsequent iterations of the outer loop. Therefore the cout will never be executed again.

In the code you commented out, j gets reset (reinitialized) to 0 for every iteration of the outer loop and therefore it will print all rows.

Answer 2

In the first one, when vvi[0] is done, j is 3 .
When it goes to vvi[1] j is still 3 so j < n is false.

In the second one, when the outer loop starts the second iteration, j will be set to 0 again.

This code is equivalent with the second one(in the result):

    for(; i < m; ++i) {
        for(j = 0; j< n; ++j) {
            cout << vvi[i][j];
        }
    }