36.Storing elements into an array using a "for" loop

 Storing elements into an array using a "for" loop:

A for loop is used to accept elements from keyboard and stores into the array.

Here the "for" loop increments the value of "i" by 1 for every iteration from 0 to n-1 where "n" is the size of array.

For every iteration the scanf() accepts an element and stores into a[i].

Here elements may be inserted one by one or all at once by separating with spaces. 

 

Accessing the elements of array using "for" loop:

A for loop is used to access the elements of array one after another.

Here the "for" loop increments the value of "i" by 1 for every iteration from 0 to n-1 where "n" is the size of array.

For every iteration an element may be fetched (accessed) through a[i].

Specification1:

Accept 5 elements from the keyboard and print them in accepted order.

Program:

#include<stdio .h="">

void main()

{

int i,a[5];

printf("Enter 5 integers:\n");

for(i=0;i<5 22="" 45="" 5="" 76="" 77="" 98="" a="" accept="" accepted="" an="" and="" are:="" array="" d="" elements="" enter="" execution:="" for="" he="" i="" in="" include="" integers:="" into="" n="" order.="" print="" printf="" program:="" scanf="" specification2:="" stdio.h="" the="" them="" vector="">

void main()

{

int n,i,a[50];

printf("How many elements:");

scanf("%d",&amp;n);

printf("Enter %d elements:\n",n);

for(i=0;i<n 0="" 12="" 34="" 45="" 4="" 50="" 56="" 65="" 67="" 6="" 78="" 87="" a="" accept="" accessed="" allocates="" allocation="" an="" and="" are:="" array="" but="" d="" elements:="" elements="" enter="" example="" execution1:="" execution2:="" explained:="" for="" from="" given="" he="" how="" i="" in="" include="" into="" is="" logic:="" many="" memory="" n-1.="" n="" of="" only="" order.="" print="" printf="" program:="" reverse="" scanf="" specifiation2:="" stdio.h="" that="" the="" them="" to="" use="" user.="" we="">

void main()

{

int n,i,a[50];

printf("How many elements?");

scanf("%d",&amp;n);

printf("Enter %d elements:\n",n);

for(i=0;i<n a="" d="" for="" i="n-1;i" in="" lements="" n="" order:="" printf="" reverse="" scanf="">=0;i--)

  printf("%5d",a[i]);

}

Execution:

How many elements? 5

Enter 5 elements:

12 67 54 34 98

Elements in reverse order:

98 34 54 67 12

Specification3:

Accept "n" elements into a vector and print the sum of all the elements.

Logic:

Program:

#include<stdio .h="">

void main()

{

int n,i,a[50],sum;

printf("How many elements?");

scanf("%d",&amp;n);

printf("Enter %d elements:\n",n);

for(i=0;i<n 12="" 265="" 34="" 54="" 5="" 67="" 98="" a="" accept="" all="" an="" and="" array="" d="" elements.="" elements:="" elements="" enter="" even="" execution:="" for="" how="" i="" include="" into="" logic:="" many="" n="" odd="" of="" print="" printf="" program:="" scanf="" specification4:="" stdio.h="" sum="" the="" um="">

void main()

{

int n,i,a[50],esum,osum;

printf("How many elements?");

scanf("%d",&amp;n);

printf("Enter %d elements:\n",n);

for(i=0;i</n></stdio></n></n></5></stdio>