C:只有if语句中的第一个条件有效

Question

我目前正在 PuTTY 中编码一些内容，并且我的 if 语句似乎遇到了技术困难。我的程序目前还不完整，但我想在完成代码之前修复这个错误。该程序将编译良好并打印；然而，只有进行第一种类型的微分才有效，而尝试进行积分或其他类型的微分则会退出程序:

int main()
{
    int probtype, diftype, intype,  vartype, vartest;
    probtype = 0;
    diftype = 0;
    intype = 0;
    vartype = 0;
    vartest = 0;
    char variable, variable2;
    float  fpower, fconstant, fpower2, fconstant2, fdivisor;
    int ipower, iconstant, ipower2, iconstant2, idivisor;
    printf("Alright, starting off would you like to do differentiation or integration? Type '1' for differentiation or '2' for integration. ");
    scanf("%d", &probtype);
    if (probtype == 1)
    {
        printf("Okay, what kind of differentiation problem are you interested in? Type 1 for 'Power Rule with a Constant', Type 2 for 'Product Rule', Type 3 for 'Quotient Rule, Type 4 for 'Chain Rule', Type 5 for 'Trigonometric Problems', Type 6 for 'Exponential Problems' or Type 7 for 'Natural Log Problems. ");
        scanf("%d", &diftype);
        if (diftype == 1)
        {
            printf("Before we begin. Type 1, if the constant and power are both integers, type 2 if the constant is a decimal and the power is an integer, and type 3 for all other cases. " );
            scanf("%d", &vartype);
            if (vartype == 1)
            {
                printf("First off enter the function's constant. If it has no visible constant enter 1. ");
                scanf("%d", &iconstant);
                printf("Alright, next enter the variable you are using. ");
                scanf("%s", &variable);
                printf("Finally, enter the power of the variable. ");
                scanf("%d", &ipower);
                iconstant = iconstant * ipower;
                ipower = ipower - 1;
                if (ipower == 0)
                {
                    printf("The derivative of the function would be: %d \n", iconstant);
                }
                else
                {
                    printf("The derivative of the function would be: %d%s^%d \n", iconstant, &variable, ipower);
                }
            }
            if (vartype == 2)
            {
                printf("First off enter the function's constant. Enter it in DECIMAL form! ");
                scanf("%f", &fconstant);
                printf("Next, enter the variable you are using. ");
                scanf("%s", &variable);
                printf("Finally, enter the power of the variable. ");
                scanf("%d", &ipower);
                fconstant = fconstant * ipower;
                ipower = ipower - 1;
                if (ipower == 0)
                {
                    printf("The derivative of the function would be %f \n", fconstant);
                }
                else
                {
                    printf("The derivative of the function would be %f%s^%d \n", fconstant, &variable, ipower);
                }
            }
            if (vartype == 3)
            {
                printf("First off, enter the function's constant. Enter it in DECIMAL form! ");
                scanf("%f", &fconstant);
                printf("Okay, now enter which variable you are using. ");
                scanf("%s", &variable);
                printf("Finally, enter the power of the variable. Enter this in DECIMAL form also! ");
                scanf("%f", &fpower);
                fconstant = fconstant * fpower;
                fpower = fpower - 1;
                printf("The derivative of the function would be: %f%s^%f \n", fconstant, &variable, fpower);
            }
            else if (diftype == 2)
            {
                printf("Enter the first function's constant. If it doesn't have one, type a 1 ");
            }
            else if (diftype == 3)
            {
                printf("Quotient Rule");
            }
            else if (diftype == 4)
            {
                printf("Chain Rule");
            }
            else if (diftype == 5)
            {
                printf("Trigoometric Problems");
            }
            else if (diftype == 6)
            {
                printf("Exponential Problems");
            }
            else if (diftype == 7)
            {
                printf("Natural Log Problems");
            }
        }
        else if (probtype == 2)
        {
            printf("Okay, now what kind of integration problem are you interested in? Type 1 for 'Indefinite Integrals', Type 2 for 'Definite Integrals', Type 3 for 'Substitution', Type 4 for 'Trignometric Integrals', Type 5 for 'Integrations by Part', Type 6 for 'Exponential Problems', Type 7 for 'Natural Log Problems. ");
            scanf("%d", &intype);
            if (intype == 1)
            {
                printf("Before we get started: type 1 if both the constant and power are integers, type 2 if the constant is an integer and the power is a decimal, type 3 if the constant is a decimal and the power is an integer, and type 4 if both are decimals." );
                scanf("%d", &vartype);
                if (vartype == 1)
                {
                    printf("Alright first off, enter the constant. If there is no constant, type a 1. " );
                    scanf("%d", &iconstant);
                    printf("Is there a variable in this equation? Type 1 if there is, Type 2 if there isn't. ");
                    if (vartest == 2)
                    {
                        printf("The indefinite integral of the function is: %dx + C", iconstant);
                    }
                    else
                    {
                        printf("Next, enter the variable that will be used for the problem. ");
                        scanf("%s", &variable);
                        printf("Finally enter the power of the function. ");
                        scanf("%d", &ipower);
                        ipower = ipower + 1;
                        idivisor = ipower + 1;
                        printf("The indefinte integral of the function is:  %d%s^%d / %d \n", iconstant, &variable, ipower, idivisor);
                    }
                }
            }
            else if (intype == 2)
            {
                printf("Definite Integrals");
            }
            else if (intype == 3)
            {
                printf("Substitution");
            }
            else if (intype == 4)
            {
                printf("Trignometric Integrals");
            }
            else if (intype == 5)
            {
                printf("Integrations by Part");
            }
            else if (intype == 6)
            {
                printf("Exponential Problems");
            }
            else if (intype == 7)
            {
                printf("Natural Log Problems");
            }
        }
    }
}

我似乎无法找到导致错误的原因，如果我的信息不够具体，我深表歉意。我知道有更有效的方法来执行此程序，因此请不要更改我的代码，而只是向我展示我做错了什么以及如何修复它。谢谢。

Answer 1

这是因为条件 else if (probtype == 2) 位于条件 if (probtype == 1) 内部，并连接到其他一些 if。相反，在之前添加另一个大括号:

} // <--- Add an extra brace here in addition to what you have.
else if (probtype == 2)
{

并删除外部 if 语句末尾的一个。

您有太多嵌套的 if else 语句。我建议您重新编写代码以使用 switch-case 和/或对子任务使用单独的函数，以便您的代码更具可读性并且您能够轻松调试。