To break nested loop

Question

I want to break the inner loop and increase the value of x if I find something at Matrix[x][y] but if I break in the y for loop it will just increase the value at y loop and not in the x loop.

int matrix[5][5]; //let's assume there are values inside
for(int i = 0; i < 5; i++)
   {
     for(int j = 0; j<5;j++)
      {
        int radius = 1;
        for(int x = radius - 1; x <= radius + i; x ++)
         {
          for(int y = radius -1; y <= radius +j; y++)
            {
               if((x+i) >= 0 && (y+j)>=0)
                 {
      //for example I find the number 45 in the matrix, then the search radius should be increased by 1 and start it over again.
                  if(matrix[x][y] == 45)
                     {
                       radius++;
                       break; // but if i break here the x value wont be changed
                      }

                  }
             }
          }
      }

You could use a variable for setting x and then increase it by 1 when found. — Ryan Wilson, Mar 15 '18 at 12:17
@Newuser1234567 Why did you decide that x will not be incremeneted after this statement break; // but if i break here the x value wont be changed? — Vlad from Moscow, Mar 15 '18 at 12:18
Yeah i want to continue the search (like a picture pixel search in openCv )and increase the radius where i stopped it @king_nak — Newuser1234567, Mar 15 '18 at 12:20
I would use a `goto` here. There is really no other way of breaking _partially_ out of a nested loop without adding checking code to the level you want to break to. (Boolean flags, numerical checks, etc.) — Arnav Borborah, Mar 15 '18 at 12:20
@VladfromMoscow I wrote a program with 10 x 10 matrix and sum the neighours and how x and y goes and somehow the x didnt changed only the y value — Newuser1234567, Mar 15 '18 at 12:21
Can you refactor the code? Might be easier to do if you could stick a `return` somewhere. — StoryTeller - Unslander Monica, Mar 15 '18 at 12:21
What happens to the value of 'j' and 'i'? If you increment the radius, it will break the 'y' loop, and then test for 'x' when i=0 will cause that loop to end. If 'j' is incremented, you know that's your problem. — HungryBeagle, Mar 15 '18 at 12:21
also named loop idiom: https://en.wikibooks.org/wiki/More_C%2B%2B_Idioms/Named_Loop — DaBler, Mar 15 '18 at 12:29

score 0 · Answer 1 · answered Mar 15 '18 at 12:24

0

Of non-local exit mechanisms C++ provides throwing:

try {
    for(int x = 0; x < 10; ++x) {
        for(int y = 0; y < 3; ++y) {
            std::cout << x << ", " << y << std::endl;
            if(y == 1 && x == 3) throw true;
        }
    }
}
catch(bool) {}
std::cout << std::endl;

Additionally you can modify the outer loop's variable so the final condition holds:

for(int x = 0; x < 10; ++x) {
    for(int y = 0; y < 3; ++y) {
        std::cout << x << ", " << y << std::endl;
        if(y == 1 && x == 3) {
            x = 10;
            break;
        }
    }
}
std::cout << std::endl;

(Or, if you don't want to modify the variable itself, say, it's not local to the loop and you'll need its value afterwards, add one more flag to the loop condtion:

bool continueOuter{true};
for(int x = 0; continueOuter && x < 10; ++x) {
    for(int y = 0; y < 3; ++y) {
        std::cout << x << ", " << y << std::endl;
        if(y == 1 && x == 3) {
            continueOuter = false;
            break;
        }
    }
}

)

answered Mar 15 '18 at 12:24

bipll

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use of exceptions would be inappropriate for this case. Needing to increase the radius of the search is not an "exceptional circumstance". It's an expected part of the algorithm. – Richard Hodges Mar 15 '18 at 12:26
And where exactly do I use exceptions? – bipll Mar 15 '18 at 12:27
2

try/catch/throw involve exceptions whether or not you derive the actual exception's type from std::exception. `throw int(1)` throws an exception of type `int`. – Richard Hodges Mar 15 '18 at 12:29
Can you kindly clarify: by "throws an exception" do you mean it is actually an exception from something, or is it a colloquial term for a non-local exit? – bipll Mar 15 '18 at 13:10
1

see [expr.throw] in the c++ standard draft: "Evaluating a throw-expression with an operand throws an exception (18.1); the type of the exception object is determined by removing any top-level cv-qualifiers from the static type of the operand and adjusting the type from “array of T” or function type T to “pointer to T”" http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2017/n4659.pdf – Richard Hodges Mar 15 '18 at 13:15

To break nested loop

1 Answers1