Thank you for your assistance in this matter. I appreciate your help
This program does exactly what you're saying
#include<stdio.h>
int main()
{
int array[5] = { 5, 6, 7, 8, 9};
int i = 0;
printf("%d, %d, %d, %d, %d", array[0], array[1], array[2], array[3], array[4]);
return 0;
}But you are losing site of the big picture -- we are exploring using arrays to efficiently solve problems and we are working with "toy" problems to keep things simple. So while we might be working with five values in our little programs here, we are trying to develop skills and techniques that allow us to write programs that solve similar problems but that are MUCH larger in scale.
Check it this program
#include<stdio.h>
int main (void)
{
int i, j, k, loop, number, size;
int sum = 0;
int array[20];
printf ( "Array Size : " ) ;
scanf ( "%d", &size ) ; /* Get the size of array */
if ( size > 0 && size < 20 ) /* Check valid size */
{
/* Loop to repeat process two times */
for ( loop = 0; loop < 2; loop++ )
{
printf ( "\nEnter array Elements \n" ) ;
for ( i = 0; i < size ; i++ )
{
scanf ( "%d", &array[i] ) ; /* Store array elements */
}
printf ( "value : ", array[i] ); /*Print list of values on one line */
for ( j = 0; j < size; j++ )
{
printf ( " %d ", array[j] ) ;
}
for( k = 0; k < 5; k++ )
{
sum = sum + array[k] ;
}
printf ( "\nsum : %d ", sum ); /* Print out the sum of the five values */
sum = 0 ;
printf ( "\nEnter number : " ) ; /*Ask the user which value they want to double, */
scanf ( "%d", &number ) ;
{
if ( ( 0 <= number ) && ( number < 5 ) )
{
printf ( "Double value : %d \n", array[number] * 2 ) ;
}
else
{
printf ( "invalid number \n" ) ;
}
}
}
}
else
{
printf ( "Array size is not acceptable" ) ;
}
return 0;
}#include<stdio.h>
int main (void)
{
int i, j, k, loop, number, size;
int sum = 0;
int array[20];
printf ( "Array Size : " ) ;
scanf ( "%d", &size ) ; /* Get the size of array */
if ( size > 0 && size < 20 ) /* Check valid size */
{
/* Loop to repeat process two times */
for ( loop = 0; loop < 2; loop++ )
{
printf ( "\nEnter array Elements \n" ) ;
for ( i = 0; i < size ; i++ )
scanf ( "%d", &array[i] ) ; /* Store array elements */
printf ( "value : ", array[i] ); /*Print list of values on one line */
for ( j = 0; j < size; j++ )
printf ( " %d ", array[j] ) ;
for( k = 0; k < 5; k++ )
sum = sum + array[k] ;
printf ( "\nsum : %d ", sum ); /* Print out the sum of the five values */
sum = 0 ;
printf ( "\nEnter number : " ) ; /*Ask the user which value they want to double, */
scanf ( "%d", &number ) ;
if ( ( 0 <= number ) && ( number < 5 ) )
printf ( "Double value : %d \n", array[number] * 2 ) ;
else
printf ( "invalid number \n" ) ;
}
}
else
printf ( "Array size is not acceptable" ) ;
return 0;
}#include<stdio.h>
int main(void)
{
int i, loop, number, size, sum;
int array[20];
printf( "Array Size : " ) ;
scanf( "%d", &size ) ; /* Get the size of array */
if ( (size > 0) && (size < 20) ) /* Check valid size */
{
/* Loop to repeat process two times */
for ( loop = 0; loop < 2; loop++ )
{
printf( "\nEnter array Elements \n" ) ;
for ( i = 0; i < size ; i++ )
scanf( "%d", &array[i] ) ; /* Store array elements */
printf( "value : ", array[i] ); /* Print list of values on one line */
for ( i = 0; i < size; i++ )
printf ( " %d ", array[i] ) ;
sum = 0;
for ( i = 0; i < 5; i++ )
sum = sum + array[i] ;
printf( "\nsum : %d ", sum ); /* Print out the sum of the five values */
printf( "\nEnter number : " ) ; /* Ask the user which value they want to double, */
scanf( "%d", &number ) ;
if ( ( 0 <= number ) && ( number < 5 ) )
printf( "Double value : %d \n", array[number] * 2 ) ;
else
printf( "invalid number \n" ) ;
}
}
else
printf( "Array size is not acceptable" ) ;
return 0;
}I've written following program to complete all requirement
#include <stdio.h>
int PassValue, loop, size , i, j, k, sum, number, array[20];;
int main ()
{
printf ("How many pass you want to check : ");
scanf ("%d", &PassValue);
if ( PassValue > 0 && PassValue < 8 ) /* Check valid pass */
{
printf("pass : %d \n", PassValue);
}
else
{
printf("Please try again \n");
printf("Pass value you enterd is not valid \n");
printf("1. Pass value shouldn't be more then 8 \n");
printf("2. You can't enter neagtive value \n");
while (!( PassValue > 0 && PassValue < 8 ))
{
printf ("How many pass you want to check :");
scanf ("%d", &PassValue);
printf("Please try again \n");
printf("Pass value you enterd is not valid \n");
printf("1. Pass value shouldn't be more then 8 \n");
printf("2. You can't enter neagtive value \n");
if ( PassValue > 0 && PassValue < 8 )
{
printf("pass : %d \n", PassValue);
break;
}
}
}
/* Loop to repeat process n times */
for ( loop = 0; loop < PassValue; loop++ )
{
printf ("Enter array size : ");
scanf ("%d", &size);
if ( size > 0 && size < 32 ) /* Check valid size */
{
printf("Entered Array size : %d \n", size);
}
else
{
printf("Please try again \n");
printf("size you enterd is not valid \n");
printf("1. number shouldn't be more then 32 \n");
printf("2. You can't enter neagtive value \n");
while (!( size > 0 && size < 32 ))
{
printf ("Enter array size ");
scanf ("%d", &size);
printf("Please try again \n");
printf("size you enterd is not valid \n");
printf("1. number shouldn't be more then 32 \n");
printf("2. You can't enter neagtive value \n");
if(size > 0 && size < 32 )
{
printf("Entered array size : %d \n", size);
break;
}
}
}
printf ( "\nEnter array Elements \n" ) ;
for ( i = 0; i < size ; i++ )
scanf ( "%d", &array[i] ) ;
printf ( "value : ", array[i] );
for ( j = 0; j < size; j++ )
printf ( " %d ", array[j] ) ;
for( k = 0; k < size; k++ )
sum = sum + array[k] ;
printf ( "\nsum : %d ", sum ); /* Print out the sum of the five values */
sum = 0 ;
printf ("\nEnter number you want to double : ");
scanf ("%d", &number);
if ( number > 0 && number < size ) /* Check valid number */
{
printf("Enterd number : %d \n", number);
}
else
{
printf("Please try again \n");
printf("number you enterd is not valid \n");
printf("1. number shouldn't be more then size \n");
printf("2. You can't enter neagtive value \n");
while (!( number > 0 && number < size ))
{
printf ("\nEnter number you want to double :");
scanf ("%d", &number);
printf("Please try again \n");
printf("number you enterd is not valid \n");
printf("1. number shouldn't be more then size \n");
printf("2. You can't enter neagtive value \n");
if(number > 0 && number < size)
{
printf("Enterd number : %d \n", number);
break;
}
}
}
printf ( "Double value : %d \n", array[number] * 2 ) ;
}
return 0;
}I am stuck I don't have any idea how to remove last commas.
I tried to follow that style but I was getting so many errors so I used enclosing braces
I would try my best for code formatting
What in the problem specifications says anything about making multiple passes?
I tried to make universal program to complete following requirement
So let's consider this aspect of the problem in detail.
printf("%d" data[0])
for(i = 1; i < n; i++)
{
printf(", ");
printf("%d", data[i]);
}
printf("\n");for(i = 0; i < n; i++)
{
printf("%d", data[i]);
if ( i != n - 1 )
printf(", ");
}
printf("\n");for(i = 0; i < n; i++)
{
if ( i != 0 )
printf(", ");
printf("%d", data[i]);
}
printf("\n");for(i = 0; i < n; i++)
{
printf("%d", data[i]);
if ( i == n - 1 )
printf("\n");
else
printf(", ");
}for(i = 0; i < n; i++)
printf("%d%s", data[i], (i == n-1)? "\n" : ", ", data[i]);That's fine. It's a valid approach, but you do need to be able to read and write code that doesn't enclose single statements, both because you will see a lot of code that uses that style and because, if you can't, that means that you really don't know program structure and syntax very well. But it isn't something that has to happen overnight.
If you want to solve a different problem, fine. But notice how you posted the program requirements as I stated them, then claimed that you had written a program that satisfied those requirements, the posted code that solved some other unstated set of requirements and NOT the program requirements as posted. That's a recipe for nothing but confusion.
I am stuck in step five. How to replace double value with original value
#include<stdio.h>
int main ()
{
int n = 5, sum = 0;
int i, j, k, number, array[n];
printf ("Enter Value \n");
for (i = 0; i < n ; i++)
scanf("%d", &array[i]);
for(i = 0; i < n; i++)
{
printf("%d", array[i]);
if ( i == 5 - 1 )
printf("\n");
else
printf(", ");
}
for (k = 0; k < n ; k++)
sum = sum + array[k];
printf ("sum : %d", sum);
printf ("\nEnter number \n");
scanf ("%d", &number);
if (number >0 && number <5)
printf ("Double value : %d \n", array[number-1] * 2 );
return 0;
}Here is program for single variable
#include<stdio.h>
int main ()
{
int x, Double;
printf ("\nEnter value \n");
scanf ("%d", &x);
printf ("x = %d \n", x);
printf ("x = %d", x * 2);
return 0;
}You are not doing what was specified.
#include<stdio.h>
int main ()
{
int x, Double;
printf ("\nEnter value \n");
scanf ("%d", &x);
printf ("x = %d \n", x);
/* Put in one line of code here that CHANGES the
value that is stored in x so that it is twice the
value that was originally stored in x. */
printf ("x = %d \n", x);
return 0;
}I hope this is what you are asking
#include<stdio.h>
int main ()
{
int x;
printf ("\nEnter value \n");
scanf ("%d", &x);
printf ("x = %d \n", x);
x = x * 2;
printf ("x = %d \n", x);
return 0;
}I am trying. but It is difficult to replace array value with a new value. I have been trying many time but I couldn't complete it
Show your attempt -- I'm still not very good at reading minds.
This is my attempt
#include<stdio.h>
int main ()
{
int n = 5, sum = 0;
int i, j, k, number, array[n];
printf ("Enter Value \n");
for (i = 0; i < n ; i++)
scanf("%d", &array[i]);
for(i = 0; i < n; i++)
{
printf("%d", array[i]);
if ( i == 5 - 1 )
printf("\n");
else
printf(", ");
}
for (k = 0; k < n ; k++)
sum = sum + array[k];
printf ("sum : %d", sum);
printf ("\nEnter number \n");
scanf ("%d", &number);
if (number >0 && number <5)
array[number-1] = array[number-1] * 2;
printf ("Double value : %d \n", array[number-1]);
array[j] = array[number-1];
for (j = 0; j < n ; j++)
printf(" value : %d ", array[j]);
return 0;
}
