How to break out of nested loops?

问题

If I use a break statement, it will only break inner loop and I need to use some flag to break the outer loop. But if there are many nested loops, the code will not look good.

Is there any other way to break all of the loops? (Please don\'t use goto stmt.)

for(int i = 0; i < 1000; i++) {
   for(int j = 0; j < 1000; j++) {
       if(condition) {
            // both of the loops need to break and control will go to stmt2
       }
   }

}

stmt2

回答1:

Use:

if (condition) {
    i = j = 1000;
    break;
}

回答2:

No, don't spoil the fun with a break. This is the last remaining valid use of goto ;)

If not this then you could use flags to break out of deep nested loops.

Another approach to breaking out of a nested loop is to factor out both loops into a separate function, and return from that function when you want to exit.

Summarized - to break out of nested loops:

use goto
use flags
factor out loops into separate function calls

Couldn't resist including xkcd here :)

source

Goto's are considered harmful but as many people in the comments suggest it need not be. If used judiciously it can be a great tool. Anything used in moderation is fun.

回答3:

bool stop = false;
for (int i = 0; (i < 1000) && !stop; i++)
{
    for (int j = 0; (j < 1000) && !stop; j++)
    {
        if (condition)
            stop = true;
    }
}

回答4:

One way is to put all the nested loops into a function and return from the inner most loop incase of a need to break out of all loops.

function() 
{    
  for(int i=0; i<1000; i++)
  {
   for(int j=0; j<1000;j++)
   {
      if (condition)
        return;
   }
  }    
}

回答5:

I think goto will solve the problem

for(int i = 0; i < 1000; i++) {
    for(int j = 0; j < 1000; i++) {
        if (condition) {
            goto end;
        }
    }
}

end:
stmt2

回答6:

You'll need a boolean variable, if you want it readable:

bool broke = false;
for(int i = 0; i < 1000; i++) {
  for(int j = 0; j < 1000; i++) {
    if (condition) {
      broke = true;
      break;
    }
  }
  if (broke)
    break;
}

If you want it less readable you can join the boolean evaluation:

bool broke = false;
for(int i = 0; i < 1000 && !broke; i++) {
  for(int j = 0; j < 1000; i++) {
    if (condition) {
      broke = true;
      break;
    }
  }
}

As an ultimate way you can invalidate the initial loop:

for(int i = 0; i < size; i++) {
  for(int j = 0; j < 1000; i++) {
    if (condition) {
      i = size;
      break;
    }
  }
}

回答7:

Use this wise advice from LLVM team:

"Turn Predicate Loops into Predicate Functions"

See:

http://llvm.org/docs/CodingStandards.html#turn-predicate-loops-into-predicate-functions

回答8:

If you need the values of i and j, this should work but with less performance than others

for(i;i< 1000; i++){    
    for(j; j< 1000; j++){
        if(condition)
            break;
    }
    if(condition) //the same condition
        break;
}

回答9:

Caution: This answer shows a truly obscure construction.

If you are using GCC, check out this library. Like in PHP, break can accept the number of nested loops you want to exit. You can write something like this:

for(int i = 0; i < 1000; i++) {
   for(int j = 0; j < 1000; j++) {
       if(condition) {
            // break two nested enclosing loops
            break(2);
       }
   }
}

回答10:

for(int i = 0; i < 1000; i++) {
   for(int j = 0; j < 1000; i++) {
       if(condition) {
            goto end;
   }
} 

end:

回答11:

int i = 0, j= 0;

for(i;i< 1000; i++){    
    for(j; j< 1000; j++){
        if(condition){
            i = j = 1001;
            break;
        }
    }
}

Will break both the loops.

回答12:

i = 0;

do
{
  for (int j = 0; j < 1000; j++) // by the way, your code uses i++ here!
  {
     if (condition)
     {
       break;
     }
  }

  ++i;

} while ((i < 1000) && !condition);

回答13:

for(int i = 0; i < 1000; i++) {
    for(int j = 0; j < 1000; i++) {
       if(condition) {
          func(para1, para2...);
          return;
       }
    }
}

func(para1, para2...) {
    stmt2;
}

来源：https://stackoverflow.com/questions/9695902/how-to-break-out-of-nested-loops

标签

nested-loops

break