Question

I am writing some C code and in my code I have two nested loops. On a particular condition I want to break out of the inner loop and continue the outer loop. I tried to achieve this using a label at the end on the outer loop’s code and, on the condition, goto that label. However gcc gives an error that I cannot have a label at the end of a compound statement. Why not?

Note 1: This is not a switch statement and that question has been answered elsewhere.

Note 2: This is not a question about style and whether I should or should not be using goto statements or conditional variables instead.

EDIT: People have asked for an example and I can give a slightly facile example of checking if an array is a subarray of another array

    int superArray[SUPER_SIZE] = {...}, subArray[SUB_SIZE] = {...};
    int superIndex, subIndex;

    for (superIndex=0; superIndex<SUPER_SIZE-SUB_SIZE; superIndex+=1)
    {
      for (subIndex=0; subIndex<SUB_SIZE; subIndex+=1)
        if (superArray[superIndex+subIndex] != subArray[subIndex])
          goto break_then_continue;

      // code that executes if subArray is a sub array

      break_then_continue:
    }

Answer 1

In the standard it’s explicitly said that labels belong to a statement, therefore a simple semicolon (;) after your label can circumvent the problem you are running in to, since that counts as a statement.

There is even an example of the use of an “empty“¹ statement in 6.8.3/6.

EXAMPLE 3 A null statement may also be used to carry a label just
before the closing } of a compound statement

while (loop1) {
  /* ... */

  while (loop2) {
    /* ... */

    if (want_out)
      goto end_loop1;

    /* ... */
  }

  /* ... */

  end_loop1: ;
}

¹ In the standard this is referred to as a null statement.

---

6.8.1 Labeled statements

Syntax
  1 labeled-statement:
      identifier : statement
      case constant-expression : statement
      default : statement

Notice that statement isn’t optional in the above quotation.

---

• open-std.org: n1124.pdf