Programming Questions

A function cannot be defined inside the another function, but a function can be called inside a another function.

Error occurs because we have mixed the ANSI prototype with K & R style of function definition.

When we use ANSI prototype for a function and pass a float to the function it is promoted to a double. When the function accepts this double into a float a type mismatch occurs hence the error.

The remedy for this error could be to define the function as :

int fun (float aa)

{

  ....

}

00 00 AC 40

ptr pointer is constant. In ptr = yourbuf the program is trying to modify it, hence an error.

#include<stdio.h>
int main(){

    int a[10],i,n,m,c=0;

    printf("Enter the size of an array: ");
    scanf("%d",&n);

    printf("Enter the elements of the array: ");
    for(i=0;i<=n-1;i++){
         scanf("%d",&a[i]);
    }

    printf("Enter the number to be search: ");
    scanf("%d",&m);
    for(i=0;i<=n-1;i++){
         if(a[i]==m){
             c=1;
             break;
         }
    }
    if(c==0)
         printf("The number is not in the list");
    else
         printf("The number is found");

    return 0;
}

Sample output:
Enter the size of an array: 5
Enter the elements of the array: 4 6 8 0 3
Enter the number to be search: 0
The number is found

atoi()         Converts a string to an integer.

itoa()         Convert an integer to a string

gcvt()        Converts a floating-point number to a string

#include "stdlib.h"

main()

{

      char s[] = "12345";

      char buffer [15], string[20];

      int i;

      i = atoi (s);

      printf("\n%d",i);

       gcvt (20.141672, 4, buffer);

       printf ("\n%s", buffer);

       itoa(15, string,2);

        printf ("\n%s", string);

}

Rounding off a value means replacing it by a nearest value that is approximately equal or smaller or greater to the given number.

y = (int)(x + 0.5); here x is any float value. To roundoff, we have to typecast the value of x by using (int)

Example:

#include

int main ()

{

  float x = 2.6;

  int y = (int)(x + 0.5);

  printf ("Result = %d\n", y );

  return 0;

}

Result : 3