Is using goto a legitimate way to break out of two

问题:

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.

回答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;
}

回答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..

回答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.

回答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.

回答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.

回答6:

I just found this on the "Related" sidebar. An interesting thread overall, but in particular, this is an answer to my question.

回答7:

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!