while loop has terminating sign still works well

Question

I was implementing Newton Raphson method in C. The code work well. There is no error in the code.

#include<stdio.h>
#include<math.h>
#define  f(x)(x * sin(x)+cos(x))
#define df(x)(x*cos(x))
int main()
{
   float x,h,e;
   e=0.0001;
   printf("Enter the initial value of x:\n");
   scanf("%f",&x);
 do
  {
     h=-f(x)/df(x);
     x=x+h;
  }
  while(fabs(h)>e);
  printf("The value of the root is=%f",x);
  return(0);
 }
/*
Output:
Enter the initial value of x: 3
The value of the root is = 2.798386

However, I was surprised I mean how did this code work? As per c rule while statement does not have any terminating semicolon. However, in my code while(fabs(h)>e); has a semicolon yet it run well.

Can anyone tells me how does it work?

Also, "per c rule" `while(x);` is a valid statement. It would be equivalent to `while(x) {}`. — shrewmouse, Sep 20 '20 at 16:55

Roberto Bressani · Accepted Answer · 2020-09-20T17:07:56.377

What you mean is putting

while(...);
{
//some code
}

that will be interpreted as

while(...){
   //looping without any instruction (probably an infinite loop)
}
{
//some code that will be executed once if the loop exits
}

do-while loop has the code executed before the condition (so at least once differently from simple while loop). The correct syntax has a semicolumn:

do{
   //your code to be executed at least once
}while(...);

score 0 · Answer 2 · answered Sep 20 '20 at 16:52

0

So the answer to your question is that :

 do
  {
     h=-f(x)/df(x);
     x=x+h;
  }
  while(fabs(h)>e);

is not a while statement, it's a do-while statement.

answered Sep 20 '20 at 16:52

shrewmouse

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while loop has terminating sign still works well

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