c++ - 使用 goto 是打破两个循环的合法方式吗？

Question

我正在解决 Project Euler 上的问题 9 .在我的解决方案中，我使用“goto”语句来打破两个 for 循环。问题如下:

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.

我的解决方案是 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;

由于“goto”语句在 c++ 程序员中不是很流行，我想知道这是否可以被认为是“goto”的合理使用。或者，对于不需要“goto”的问题，是否有更好的解决方案。我的意思不是一个解决方案，它只是避免“goto”，而是以改进算法的方式避免“goto”。

Answer 1

return 是一个“结构化”goto，许多程序员觉得它更容易接受!所以:

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;
}