Infinite loop.

[Ubuntu2014]

Hello. I made a C program and used while(1) to create loop and defined condition is if(a<0)break; . But when I by mistake typed an alphabet, it continued as an infinite loop. I searched Google but couldn't understand. Please tell me how do I cure this ? If anyone has time, may I have one example ?

[grumpy]

We're not mindreaders, you know.

Try providing the smallest possible sample of code you can that exhibits your main problem.

I can make an informed guess, but you need to learn how to ask questions that don't waste people's time unnecessarily.

[Ubuntu2014]

#include<stdio.h> 
#include<math.h> 
int main() 
{ 
double zeta5 = 24.88626612344087823195277167496882003336994206804590748738062 , a , b , c , d ,  
e = 2.718281828459045235360287471352662497757247093699959574966967 , f , term , integral , p , v; 
 
printf("\nHello.") 
 
while(1) 
    { 
        while(1) 
            { 
 
            printf("\nEnter the upper limit (>= 0) : ");         
            scanf("%lf",&p) ; 
            if(p<0)printf("please enter a whole number !\n"); 
            else break; 
 
            }
              .
              .
              .

.
.
.

[CodeMonkey]

#include<stdio.h> 
#include<math.h> 

int main() 
{ 
    const double zeta5 = 24.88626612344087823196;
    double a, b, c, d;   
    const double e = 2.718281828459045235360287;
    double f, term, integral, p, v; 
 
    printf("\nHello.") 
 
    while(1) 
    {  
        while(1) 
        { 
            printf("\nEnter the upper limit (>= 0) : ");         
            scanf("%lf",&p) ; 
            if (p < 0)
                printf("please enter a whole number !\n"); 
            else 
                break; 
        }
        /* . . .
            The rest of the outer loop
            . . . 
        */
    }
    /* . . .
        After the outer loop
        . . .
    */
}

What were your inputs for the given code, and what problem did it give you?

[grumpy]

The problem is obvious.

scanf("%lf", &p) will return if it encounters a character (such as 'X') that would not be expected when reading a floating point value. It will not modify p AND it will leave the character in the stream. The next call of scanf() will encounter the character, so will return immediately leaving p unmodified and the character in the stream. And so on.

Check the return value of scanf() to see if it successfully read the required value. Your choices - if it hasn't - are to either back out of the loop (as long as the outer loops don't rely on a value of p being read) or to read a single character from the stream (e.g. using "scanf("%c", &some_character)") and then try to read the floating point value again.


I won't even want to ask why you're describing a non-negative value as a "whole number".

[Ubuntu2014]

#include<stdio.h> 
#include<math.h> 

int main() 
{ 
    const double zeta5 = 24.88626612344087823196;
    double a, b, c, d;   
    const double e = 2.718281828459045235360287;
    double f, term, integral, p, v; 
 
    printf("\nHello.") 
 
    while(1) 
    {  
        while(1) 
        { 
            printf("\nEnter the upper limit (>= 0) : ");         
            scanf("%lf",&p) ; 
            if (p < 0)
                printf("please enter a whole number !\n"); 
            else 
                break; 
        }
        /* . . .
            The rest of the outer loop
            . . . 
        */
    }
    /* . . .
        After the outer loop
        . . .
    */
}

What were your inputs for the given code, and what problem did it give you?

When I typed a variable, it kept me asking whether I want to quit the program and the program did not terminate even if I typed N (for no).

[Ubuntu2014]

Is there any command in C language which goes like this :
if(a=real number)continue;
else break;

[grumpy]

Is there any command in C language which goes like this :
if(a=real number)continue;
else break;

Nope. If you look at my previous post, you'll see an explanation for your problem, and some possible solutions.

[marque]

Hello. I made a C program and used while(1) to create loop and defined condition is if(a<0)break; . But when I by mistake typed an alphabet, it continued as an infinite loop. I searched Google but couldn't understand. Please tell me how do I cure this ? If anyone has time, may I have one example ?

use atoi() standard function.

[grumpy]

use atoi() standard function.

atoi() is insufficient.

atoi("xxc") will return zero, true. So will atoi("0").

Furthermore, atoi("22x") will return 22, not zero.