Question

我有以下C代码。计算 simple_root 的语句出现两次。我想知道是否有更好的方法来组织代码，使该语句仅出现一次。

如果将 simple_root 放在最后一个之后，则可能会出现错误，例如前两个条件不匹配。如果我错了，请纠正我。

if(fabs(quartic_polynomial(a, b, c, d, e, secDerivRoot_1)) < 1e-6)
{
    triple_root = secDerivRoot_1;
    simple_root = -b/a - 3*triple_root;
}
else if (fabs(quartic_polynomial(a, b, c, d, e, secDerivRoot_2)) < 1e-6)
{
    triple_root = secDerivRoot_2;
    simple_root = -b/a - 3*triple_root;
}       
else 
    printf("Something is wrong. No common root with its 2nd derivative. \n");

Answer 1

您在评论中缺少了一些内容

如果将simple_root放在最后一个之后，则存在出错的风险，例如前两个条件不匹配

已经是这种情况-如果您不匹配这两个，则triple_root和simple_root尚未设置。我不确定在这段代码后如何使用它们，或者您的错误处理方式是什么。您可以在printf语句后立即返回，也可以将triple_root设置为NaN。

然后，您可以将simple_root公式移至if语句下方。或者，您可以将其放入函数中。单行功能可以使代码更易于理解。

Answer 2

我认为以下应该这样做。我尚未对其进行编译或测试，因此您可能需要进行一些调整。

但是您的原始措辞更具可读性，我真的发现以这种方式进行这样做没有任何优势。

使用这种方法，您总是会同时执行两个quartic_polynomial()调用，而不是有时仅执行一个调用，这可能会严重影响CPU的使用和时间，而不是复制一行代码。

我很好奇两个if语句是否都可以成立。在这种情况下您该怎么办？还是可以选择自己喜欢的人？

{
    int sec_1 = (fabs(quartic_polynomial(a, b, c, d, e, secDerivRoot_1) < 1e-6);
    int sec_2 = (fabs(quartic_polynomial(a, b, c, d, e, secDerivRoot_2) < 1e-6);

    if (! (sec_1 || sec_2)) {
        printf("Something is wrong. No common root with its 2nd derivative. \n");
    } else {
        if (sec_1)
            triple_root = secDerivRoot_1;
        else
            triple_root = secDerivRoot_2;
        simple_root = -b/a - 3*triple_root;
    } 
}

另一种方法，如果希望防止一个语句被更新，而另一个则不会，则如下所示：

#define SIMPLE_ROOT_CALC  simple_root = -b/a - 3*triple_root

if(fabs(quartic_polynomial(a, b, c, d, e, secDerivRoot_1)) < 1e-6)
{
    triple_root = secDerivRoot_1;
    SIMPLE_ROOT_CALC;
}
else if (fabs(quartic_polynomial(a, b, c, d, e, secDerivRoot_2)) < 1e-6)
{
    triple_root = secDerivRoot_2;
    SIMPLE_ROOT_CALC;
}       
else 
    printf("Something is wrong. No common root with its 2nd derivative. \n");

甚至是类似的东西

#define ROOT_CALC(x)  triple_root = (x); simple_root = -b/a - 3*triple_root;

if(fabs(quartic_polynomial(a, b, c, d, e, secDerivRoot_1)) < 1e-6)
{
    ROOT_CALC(secDerivRoot_1);
}
else if (fabs(quartic_polynomial(a, b, c, d, e, secDerivRoot_2)) < 1e-6)
{
    ROOT_CALC(secDerivRoot_2);
}       
else 
    printf("Something is wrong. No common root with its 2nd derivative. \n");

或者可能取决于与逻辑或运算符的评估短路：

{
    double x;
    if (! (((x = secDerivRoot_1), fabs(quartic_polynomial(a, b, c, d, e, secDerivRoot_1) < 1e-6)) || ((x = secDerivRoot_2), fabs(quartic_polynomial(a, b, c, d, e, secDerivRoot_2) < 1e-6) )) ) {
        printf("Something is wrong. No common root with its 2nd derivative. \n");
    } else {
        triple_root = x;
        simple_root = -b / a - 3 * triple_root;
    }
}

C：使方程式仅出现一次

2 个答案: