What is a neat way of breaking out of many for loops at once?

Question

Suppose I need to break out of three or four nested for loops at once at the occurence of some event inside the innermost loop. What is a neat way of doing that?

Answer 1

If you're using Java, you can associate labels with each for block and then reference the label after a continue statement. For example:

outerfor:
for (int i=0; i<5; i++) {
    innerfor:
    for (int j=0; j<5; j++) {
        if (i == 1 && j == 2) {
             continue outerfor;
        }
    }
}

Answer 2

If you absolutely don't want to use goto, set all loop conditions to false:

int i, j, k;

for (i = 0; i < 100; i++) {
    for (j = 0; j < 100; j++) {
        for (k = 0; k < 100; k++) {
            if (k == 50) {
                i = j = k = INT_MAX;
                break;
            }
        }
    }
}

note: a smart optimizing compiler will turn the contents of the if in a jump to the end of the outer-most loop

Answer 3

goto. This is one of the very few places where goto is the appropriate tool, and is usually the argument presented why goto isn't complete evil.

Sometimes, though, I do this:

void foo() {
    bar_t *b = make_bar();
    foo_helper(bar);
    free_bar(b);
}

void foo_helper(bar_t *b) {
    int i,j;
    for (i=0; i < imax; i++) {
        for (j=0; j < jmax; j++) {
            if (uhoh(i, j) {
                return;
            }
        }
    }
}

The idea is that I get a guaranteed free of bar, plus I get a clean two-level break out of the switch via return.

Answer 4

Dividing by 0 is the surest method I know that will break you out of any number of loops. This works because the DIV assembly instruction doesn't like such silliness.

So, you can try this:

int i, j, k;
int flag1 = 0;
int flag2 = 0;

for (i = 0; i < 100; i++) {
    for (j = 0; j < 100; j++) {
        for (k = 0; k < 100; k++) {
            if (k == 50) {
                flag1 = 1;
                flag2 = 1;
                int z = 1 / 0;  // we're outta here!!!
            }
        }
        if (flag1 == 1)break;
    }
    if (flag2 == 1)break;
}

Getting back from the trap that happens on such events left as an exercise for the reader (it's trivial).

Answer 5

One way to do it is a state machine. But i would still use goto. It's much simpler. :)

state = 0;
while( state >= 0){
    switch(state){
        case 0: i = 0; state = 1; // for i = 0
        case 1:
            i++; 
            if (i < 100)   // if for i < 100 not finished
                state = 2; // do the inner j loop
            else
                state = -1; // finish loop
        case 2: j = 0; state = 3; // for j = 0
        case 3: 
            j++;
            if (j < 100)  // if j < 100 not finished
                state = 4 // do the inner k loop
            else
                state = 1; // go backt to loop i
            break;
        case 4: k = 0; state = 5;
        case 5:
            k++;
            if (k == 50){
                state = -1;
                break;
            }
            if (k < 100) // if k loop not finished
                state = 5; // do this loop
            else
                state = 3; // go back to upper loop
            break;
        default : state = -1;
    }
}

Answer 6

If you are using GCC and this library, the break can accept the number of nested loops you want to exit:

int i, j, k;

for (i = 0; i < 100; i++) {
    for (j = 0; j < 100; j++) {
        for (k = 0; k < 100; k++) {
            if (k == 50) {
                break(3);
            }
        }
    }
}