Is using goto a legitimate way to break out of two loops? - c++

Question

I am solving problem 9 on the Project Euler. In my solution I use a "goto" statement to break out of two for loops. The Problem is the following:
A Pythagorean triplet is a set of three natural numbers, a b c, for which,
a^2 + b^2 = c^2
For example, 3^2 + 4^2 = 9 + 16 = 25 = 52.
There exists exactly one Pythagorean triplet for which a + b + c = 1000.
Find the product abc.
My solution is in c++:
int a,b,c;
const int sum = 1000;
int result = -1;
for (a = 1; a<sum; a++){
for (b = 1; b < sum; b++){
c = sum-a-b;
if (a*a+b*b == c*c){
result = a*b*c;
goto found;
}
}
}
found:
std::cout << "a:" << a << std::endl;
std::cout << "b:" << b << std::endl;
std::cout << "c:" << c << std::endl;
std::cout <<"Result:" << result << std::endl;
Since "goto" statements are not very popular among c++ programmers, i would like to know, if this could be considered a reasonable use of "goto". Or if there is a better solution for the problem that doesn't need "goto". By that I don't mean a solution which just avoids "goto", but which avoids "goto" in a way that improves the algorithm.

Answer 1

return is a "structured" goto which many programmers find more acceptable! So:
static int findit(int sum, int* pa, int* pb, int* pc)
{
for (int a = 1; a<sum; a++) {
for (int b = 1; b < sum; b++) {
int c = sum-a-b;
if (a*a+b*b == c*c) {
*pa = a; *pb = b; *pc = c;
return a*b*c;
}
}
return -1;
}
int main() {
int a, b, c;
const int sum = 1000;
int result = findit(sum, &a, &b, &c);
if (result == -1) {
std::cout << "No result!" << std::endl;
return 1;
}
std::cout << "a:" << a << std::endl;
std::cout << "b:" << b << std::endl;
std::cout << "c:" << c << std::endl;
std::cout <<"Result:" << result << std::endl;
return 0;
}

Answer 2

In my opinion it's fine to use goto in a situation like this.
Btw, the condescending preaching against goto usually comes from people who just parrot what they heard others say or read somewhere..

Answer 3

See this question about breaking out of 2 loops. There are much better answers provided than using a goto.
The best answer provided is to place your second loop into a function, and call that function from inside your first loop.
code copied from mquander's response
public bool CheckWhatever(int whateverIndex)
{
for(int j = 0; j < height; j++)
{
if(whatever[whateverIndex][j]) return false;
}
return true;
}
public void DoubleLoop()
{
for(int i = 0; i < width; i++)
{
if(!CheckWhatever(i)) break;
}
}
Though I do feel that using a goto in this case isn't quite as bad as killing kittens. But it's close.

Answer 4

I can't think of a better alternative. But one alternative not using goto would be modifying the first for-loop:
for (a = 1; a<sum && result == -1; a++){
Then break out of the second for-loop. That will work assuming the result will never be -1 after the second for-loop has been broken by break.

Answer 5

You could declare a bool found = false at the top and then add && !found to your for loop conditionals (after a < sum and b < sum) and then set found to true where your current goto is. Then make your output conditional on found being true.

Answer 6

int a,b,c,sum = 1000;
for (a = 1; a<sum; ++a)
for (b = 1; b<sum; ++b){
c = sum-a-b;
if (a*a+b*b == c*c) sum = -a*b*c;
}
printf("a: %d\n",a-1);
printf("b: %d\n",b-1);
printf("c: %d\n",c);
printf("Result: %d\n",-sum);
Also optimized result out.. :P
Anyway i love gotos!