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)

{

  ....

}

The Math.floor() function in JavaScript is used to round off the number passed as parameter to its nearest integer in Downward direction of rounding i.g towards the lesser value.

Hence, math.floor(3.6) = 3.

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