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Programming-Idioms

History of Idiom 43 > diff from v44 to v45

Edit summary for version 45 by MatthewH:
[Cpp] The comment provides an unrelated opinion that has nothing to do with the solution.
↷

Version 44

2019-09-26, 17:31:27

Version 45

2019-09-26, 17:38:12

Idiom #43 Break outer loop

Look for a negative value v in 2D integer matrix m. Print it and stop searching.

Illustration

Idiom #43 Break outer loop

Look for a negative value v in 2D integer matrix m. Print it and stop searching.

Illustration
Imports 
#include <iostream>
Imports 
#include <iostream>
Code 
auto indices = findNegativeValue (m, 10, 20);
std::cout << m[indices.first][indices.second] << '\n';

std::pair<int, int> findNegativeValue (int **m, int rows, int columns) {
  for (int i = 0; i < rows; ++i) {
    for (int j = 0; j < columns; ++j) {
      if (m[i][j] < 0) return make_pair (i, j);
    }
  }
  throw "No negative value!";
}
Code 
auto indices = findNegativeValue (m, 10, 20);
std::cout << m[indices.first][indices.second] << '\n';

std::pair<int, int> findNegativeValue (int **m, int rows, int columns) {
  for (int i = 0; i < rows; ++i) {
    for (int j = 0; j < columns; ++j) {
      if (m[i][j] < 0) return make_pair (i, j);
    }
  }
  throw "No negative value!";
}
Comments bubble
Whenever the code is as complicated as you need to break the outer loop, it is the correct time to add a new function.
Comments bubble
Whenever the code is as complicated as you need to break the outer loop, it is the correct time to add a new function.

edit: The above comment is an opinion and adds nothing to presenting a solution to the stated problem. It is a narrow-minded opinion to think such a statement can apply to every conceivable situation.