switch case exercise solution

In this post, let’s continue our previous exercise.

Let’s modify the code to handle the negative numbers.

First, let’s check the case c, which is a circle. Here, after getting the radius(r) value, if(r < 0), then you have to print “radius cannot be negative” and make the area -1. So, calculate the area only if the radius is positive. That’s why I use the else statement here. 

area=3.1415*r*r is calculated only when the radius is positive; otherwise, the area will be concluded as negative.

switch(code){
case 'c':
       printf("Circle Area calculation\n");
       printf("Enter radius(r) value:");
       scanf("%f",&r);
       if( r < 0){
              printf("radius cannot be -ve\n");
              area = -1;
       }else{
              area = 3.1415 * r * r;
       }
       break;

Handle negative numbers code

Here in the case ‘t’, we can consider two values: base and height. So, after reading the values b and h, let’s add the code. if((b<0)|| (h<0)), then print “base or height cannot be -ve”; else, it calculates the area.

case 't':
        printf("Triangle Area calculation\n");
        printf("Enter base(b) value:");
        scanf("%f",&b);
        printf("Enter height(h) value:");
        scanf("%f",&h);
        if( (b < 0) || ( h < 0)){
                    printf("base or height cannot be -ve\n");
                    area = -1;
       }else{
                     area = ( b * h)/2;
       }
       break;

Here in case ‘z’, we use 3 variables: a, b and h. So, if(((a < 0) || (b < 0) || (h < 0))), if one among these become negative, this expression will be true, then you can print “base or height cannot be negative”. That’s correct.

case 'z':
       printf("Trapezoid Area calculation\n");
       printf("Enter base1(a) value:");
       scanf("%f",&a);
       printf("Enter base2(b) value:");
       scanf("%f",&b);
       printf("Enter height(h) value:");
       scanf("%f",&h);
       if( ( (a < 0) || ( b < 0) || (h < 0) ) ){
               printf("base or height cannot be -ve\n");
               area = -1;
       }else{
               area = (( a + b)/2) * h;
       }
       break;

In case ‘s’,  if(a  <  0), then print “side cannot be negative”; Otherwise, calculate the area. 

case 's':
      printf("Square Area calculation\n");
      printf("Enter side(a) value:");
      scanf("%f",&a);
      if( a < 0){
            printf("side cannot be -ve\n");
            area = -1;
      }else{
            area = a * a;
      }
      break;

And also do the same as a rectangle. 

Let’s execute the code and see the output. First, let me try a rectangle. Enter width value, I type 4, and length value I enter negative number -5. So, it shows width or length cannot be negative. That’s correct.

Let’s try once again.  Look at Figure 2, it shows the correct output.

We completed the discussion on switch/case statements, and so we are going to use switch case statements in our embedded programming as well. That’s all about a switch/case statement.

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