问题
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?
what I do is use flags like 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;
break;
}
}
if (flag1 == 1)break;
}
if (flag2 == 1)break;
}
I don't think this is particularly neat.
How would you accomplish the same thing? (w/o using jumps)
回答1:
use goto. it's clean and simple.
回答2:
Put all the loops in a function and just return instead of break.
回答3:
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;
}
}
}回答4:
How would you accomplish the same thing? (w/o using jumps)
Why? Nothing is universally evil, and every put-upon tool has its uses (except gets()). Using goto here makes your code look cleaner, and is one of the only choices we have (assuming C). Look:
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) {
goto END;
}
}
}
}
END:
Much cleaner than all those flag variables, and it even shows more clearly what your code is doing.
回答5:
Just a wee bit better.
int i, j, k;
int flag1 = 0;
int flag2 = 0;
for (i = 0; i < 100 && !flag2; i++) {
for (j = 0; j < 100 && !flag1; j++) {
for (k = 0; k < 100; k++) {
if (k == 50) {
flag1 = 1;
flag2 = 1;
break;
}
}
}
}
But if you really need to have those loops, then it makes sense explicitly declaring in each loop what conditions must hold for it to continue, for readability.
回答6:
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.
回答7:
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
回答8:
sometimes you can use trick like this:
for (i = 0; i < 100 && !flag2; i++) {
for (j = 0; j < 100 && !flag1; j++) {
for (k = 0; k < 100; k++) {
if (k == 50) {
k = 100;
i = 100;
j = 100;
}
}
}
}
or declare addition flag in your loop:
bool end = false;
for(int i =0; i < 1000 && !end; i++) {
//do thing
end = true;
}
it costs only a line but clean, I think.
justin
回答9:
If premature completion of any cycle always means that you have to break the enclosing cycle as well, then you don't need any extra flags. The whole thing might just look as follows
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;
}
if (k < 100) break;
}
if (j < 100) break;
}
In my experience, this is what is needed in majority of cases.
回答10:
a little bit of silly self-documenting:
int i, j, k;
int done = 0;
for (i = 0; i < 100 && ! done; i++) {
for (j = 0; j < 100 && ! done; j++) {
for (k = 0; k < 100 && ! done; k++) {
if (k == 50) we_are(done);
}
}
}
//...
void we_are(int *done) {
*done = 1;
}
but really, you shouldn't have three nested for-loops. You should consider refactoring into different functions and improving your program's logic instead of doing this.
While I agree that sometimes goto really is the best solution, I think that any problem to which goto is the solution is a result of poor code.
回答11:
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).
回答12:
I'd do something like:
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) {
return;
}
}
}
}
回答13:
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);
}
}
}
}
回答14:
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;
}
}
来源：https://stackoverflow.com/questions/1586932/what-is-a-neat-way-of-breaking-out-of-many-for-loops-at-once