命令行和用户输入

Question

    #include <stdlib.h>
#include <stdio.h>

int main(int argc, char *argv[] )
{
    float out_radius,in_radius,area, perimeter, in_radius2, out_radius2;
    
    if (argc == 1) {
        printf("Enter inner radius of ring: ");
        scanf("%f",&in_radius);
        printf("Enter outer radius of ring: ");
        scanf("%f",&out_radius);
        perimeter= (2 * 3.14) * (in_radius + out_radius);
        area= 3.14 * ((in_radius * in_radius) + (out_radius * out_radius));
    }
    
    else if (argc > 1 || argc == 2) {
        in_radius2 = atof(argv[1]);
        out_radius2 = atof(argv[2]);
        perimeter = (2 * 3.14) * (in_radius2 + out_radius2);
        area = 3.14 * ((in_radius2 * in_radius2) + (out_radius2 * out_radius2));
    }
    
    printf("Area of circle: %.2f \nPerimeter of circle: %.2f\n",area,perimeter);
}

这是一个简单的程序，当给定内半径和外半径作为用户输入时计算周长和半径。我试图做到这一点，以便用户可以选择通过用户输入或命令行（无论用户决定使用哪个）输入半径。我尝试创建 and if 和 else if 语句来检查用户是否在命令行中输入数据。我的问题是当我通过命令行输入半径时，我得到 0 或有时错误的答案作为我的周长和面积。我需要做什么来解决这个问题？

编辑固定代码：

#include <stdlib.h>
#include <stdio.h>

int main(int argc, char *argv[] )
{
    float out_radius,in_radius,area, perimeter, in_radius2, out_radius2;
  
    if (argc == 3) {
        in_radius2 = atof(argv[1]);
        out_radius2 = atof(argv[2]);
        
    }
   
    else {
        printf("Enter inner radius of ring: ");
        scanf("%f",&in_radius);
        printf("Enter outer radius of ring: ");
        scanf("%f",&out_radius);
    }
    perimeter = (2 * 3.14) * (in_radius2 + out_radius2);
    area = 3.14 * ((in_radius2 * in_radius2) + (out_radius2 * out_radius2));
    
    printf("Area of circle: %.2f \nPerimeter of circle: %.2f\n",area,perimeter);
}

Answer 1

argc 是：

If the value of argc is greater than zero, the array members argv[0] through argv[argc-1] inclusive shall contain pointers to strings, ... C17dr § 5.1.2.2.1 2

If the value of argc is greater than zero, the string pointed to by argv[0] represents the program name ....

改变逻辑

if (argc == 3) {
  // Don't need to read inputs. Use argv[1], argv[2] as pointers to strings
} else {
  // Read inputs
}

嗯，OP 的更改代码 - 很难回答移动的目标。

命令行和用户输入

Command line and user input

c

command-line

user-input