C 程序:将 if 语句代码转换为不使用逻辑、关系运算符或选择结构的代码(不允许切换)

Question

我用C写下了这段代码。它相当简单明了。如何将程序末尾的 if 语句转换为不使用任何类型的逻辑或关系运算符、任何选择构造或任何数组的语句？这意味着 if 语句必须转换为简单的算术代码段。

#include<stdio.h>
#include<math.h>
int main (void)
{
    //Declarations
    int m;  //mass of the object
    int vi; //initial velocity of the object
    int ur; //coefficient of resistance
    int choice; //choice
    double net_force; //net force acting on the object
    double force_grav; //force due to gravity only
    double force_res; //force due to the resistance only
    double force_des; //Desired force in output section
    #define g 9.8
    //Input statements
    printf("Please input the mass in kilograms:");
    scanf("%d",&m);
    printf("Please input the launch speed (m/s):");
    scanf("%d",&vi);
    printf("Please input the coefficient of resistance (kg/s):");
    scanf("%d",&ur);
    //Executable Statements
    printf("Choices for calculation:n");
    printf("1. Force due to gravity onlyn");
    printf("2. Net forcen");
    printf("3. Force due to resistance onlyn");
    printf("Please enter your choice: ");
    scanf("%d", &choice);
    //Calculations
    force_grav = m * g;
    force_res = (-1) * ur * vi;
    net_force = force_res + force_grav;

    if(choice==1)
        printf("Desired force: %2.3lf", force_grav);
    else
        if(choice==2)
            printf("Desired force: %2.3lf", net_force);
        else
            printf("Desired force: %2.3lf", force_res);
   return(0);
}

Answer 1

如果这个问题应该用算术来解决，我的建议是：

force_grav = m * g;
force_res = (-1) * ur * vi;
net_force = force_res + force_grav;
force = force_grav*(2-choice)*(3-choice)/2 + 
        force_res*(1-choice)*(2-choice)/2  +
        net_force*(1-choice)*(3-choice)*(-1);
printf("Desired force: %2.3lf", force);

变量"选择"现在有效地选择，它的总和有助于项的结果。

该术语当然可以简化，但我让它"按原样"来显示这个概念。

Answer 2

这是我的第一次尝试

#include<stdio.h>
#include<math.h>
int main (void)
{
    //Declarations
    int m;  //mass of the object
    int vi; //initial velocity of the object
    int ur; //coefficient of resistance
    int choice; //choice
    double force_grav; //force due to gravity only
    double force_res; //force due to the resistance only
    const double g = 9.8;
    //Input statements
    printf("Please input the mass in kilograms:");
    scanf("%d",&m);
    printf("Please input the launch speed (m/s):");
    scanf("%d",&vi);
    printf("Please input the coefficient of resistance (kg/s):");
    scanf("%d",&ur);
    //Executable Statements
    printf("Choices for calculation:n");
    printf("1. Force due to gravity onlyn");
    printf("2. Net forcen");
    printf("3. Force due to resistance onlyn");
    printf("Please enter your choice: ");
    scanf("%d", &choice);
    //Calculations
    force_grav = m * g;
    force_res = (-1) * ur * vi;
    printf("Desired force: %2.3lfn", ((force_res*((choice&2)>>1)) + (force_grav*(choice&1))));
}

另一个可以是

printf("Desired force: %2.3lfn", ((force_res*(choice/2)) + (force_grav*(choice%2))));

小解释：

在第一个示例中，您可以使用按选择设置的位来"允许对输出求和值：

• 0x01 仅位 0 集
• 0x02 只有位 1 套
• 0x03位 0 和 2 集

在第二个示例中，您可以使用取模和整数除法来允许对值求和。

Answer 3

这是一个简单但不安全的方法。请注意，没有检查输入的值/选择，但检查需要 if...

#include<stdio.h>
#include<math.h>
int main (void)
{
    //Declarations
    int m;  //mass of the object
    int vi; //initial velocity of the object
    int ur; //coefficient of resistance
    int choice; //choice
    double net_force; //net force acting on the object
    //double force_grav; //force due to gravity only
    //double force_res; //force due to the resistance only
    //double force_des; //Desired force in output section
    double forces[3];
    #define g 9.8
    //Input statements
    printf("Please input the mass in kilograms:");
    scanf("%d",&m);
    printf("Please input the launch speed (m/s):");
    scanf("%d",&vi);
    printf("Please input the coefficient of resistance (kg/s):");
    scanf("%d",&ur);
    //Executable Statements
    printf("Choices for calculation:n");
    printf("1. Force due to gravity onlyn");
    printf("2. Net forcen");
    printf("3. Force due to resistance onlyn");
    printf("Please enter your choice: ");
    scanf("%d", &choice);
    //Calculations
    forces[0] = m * g;                   // force_grav
    forces[2] = (-1) * ur * vi;          // force_res
    forces[1] = forces[0] + forces[2];  // net_force
#if 0
    if(choice==1)
        printf("Desired force: %2.3lf", force_grav);
    else
        if(choice==2)
            printf("Desired force: %2.3lf", net_force);
        else
            printf("Desired force: %2.3lf", force_res);
#endif
   // CAUTION: SHOULD HAVE A CHECK ON THE VALUE OF CHOICE
   // E.G. if (choice > 0 && choice < 4)
   // BUT THAT'D REQUIRE AN IF......
   printf("Desired force: %2.3lfn", forces[choice-1]);
   return(0);
}