在C-中实施循环

Question

I am confused on how to complete this for loop. The mission is to read input in unix. For the input if the radius is >0 it should prompt the user each time and then if <=0 it should terminate. I am going from centimeters to square inches. My current configuration requires 2 inputs (1 prompted, 1 not) before giving output to the console. Cheers.

#include <stdio.h>
#define PI 3.14159

main()
{
float r, a;
  int y = 9999999;


  for(int i =0; i <y; i++){
     printf("Enter the circle's radius (in centimeters): ");
     scanf ("%f", &r);

        if(r>0){
           r=r;
       a = PI * r * r *2.54;

       printf("Its area is %3.2f square inches.\n", a);
   }  else {}
     }

  }

Answer 1

Your code flow is the following:

for (infinite condition) {
  scan input
  if (input > 0) {
    do things
  }
  else {
    do nothing
  }
}

So there's no way to exit out of the loop, that's why the break statement exists, to force quitting an iterative block of code:

while (true) {
  scanf ("%f", &r);
  if (r > 0) {
    // do whatever;
  }
  else
    break;
}

The break will stop the cycle when executed, just going out of the loop.

Answer 2

Consider a while loop instead:

#include <stdio.h>
#define PI 3.14159

main(){
    float r, a;

    int continueBool = 1;
    while(continueBool == 1){
         printf("Enter the circle's radius (in centimeters): ");
         scanf ("%f", &r);

         if(r>0){
              a = PI * r * r *2.54;
              //the above formula may be wrong, so consider trying:
              //a = PI * r * r/2.54/2.54;
              printf("Its area is %3.2f square inches.\n", a);
         }
         else{
             continueBool = 0;
         }
     }
}

The break statement can be dangerous if you are new to C programming, so I recommend not using it until you get a better understanding of C and break. If you do want to use break , then this could be your solution:

#include <stdio.h>
#define PI 3.14159

main(){
    float r, a;

    while(1){
         printf("Enter the circle's radius (in centimeters): ");
         scanf ("%f", &r);

         if(r<=0){
             break;
         }

         a = PI * r * r *2.54;
         //the above formula may be wrong, so consider trying:
         //a = PI * r * r/2.54/2.54;
         printf("Its area is %3.2f square inches.\n", a);
     }
}

Answer 3

r=1.0f;

// break if no. of cases exhausted or r is negative or zero
for(int i =0; i < y && r > 0; i++) 
{
     printf("Enter the circle's radius (in centimeters): ");

     if( scanf ("%f", &r) == 1) // Always check for successful scanf
     {
          a = PI * r * r/2.54/2.54; //This is correct formula

          printf("Its area is %3.2f square inches.\n", a);
     } 

 }

Answer 4

You may want to try using a while loop instead so that the question is continually prompted until the user inputs a value =>0. see if below helps (also your conversion factor was not quite right);

#include <stdio.h>
#define PI 3.14159

void main()
{
    float r, a;
    printf("Enter the cirle's radius (in centimeters):");
    scanf("%f",&r); 

    while (r>0)
    {
        a=PI*r*r*0.155; // conversion from sqcm to sqin is ~0.155
        printf("Its area is %3.2f square inches \n", a);
        printf("Enter the cirle's radius (in centimeters):");
        scanf("%f",&r);
    }
}

Answer 5

Use this:

for(int i =0; i < y; i++)
{
     printf("Enter the circle's radius (in centimeters): ");
     scanf ("%f", &r);

     if(r > 0)
     {
          a = PI * r * r *2.54;

          printf("Its area is %3.2f square inches.\n", a);
     } 
     else
     {
          break;
     }
 }