C Programming - Pointers - Discussion

#include<stdio.h>

int main()
{
int i, a[] = {2, 4, 6, 8, 10};
change(a, 5);
for(i=0; i<=4; i++)
printf("%d, ", a[i]);
return 0;
}
void change(int *b, int n)
{
int i;
for(i=0; i<n; i++)
*(b+1) = *(b+i)+5;
}


Here the entire sub function work is to change second value in the array .

The function isn't declared and defined after the main shouldn't be Wrong? Please explain.

Why? Please explain.

@All.

According to me, the processed output is;

BEFORE *(b+1) = *(b+i)+5;==>> i=0 --- *(b+1)== 4 &&&&& *(b+i) ie *(b+0)== 2,
AFTER *(b+1) = *(b+i)+5; ==>> i=0 --- *(b+1)== 7 &&&&& *(b+i) ie *(b+0)== 2,

BEFORE *(b+1) = *(b+i)+5;==>> i=1 --- *(b+1)== 7 &&&&& *(b+i) ie *(b+1)== 7,
AFTER *(b+1) = *(b+i)+5; ==>> i=1 --- *(b+1)== 12 &&&&& *(b+i) ie *(b+1)== 12,

BEFORE *(b+1) = *(b+i)+5;==>> i=2 --- *(b+1)== 12 &&&&& *(b+i) ie *(b+2)== 6,
AFTER *(b+1) = *(b+i)+5; ==>> i=2 --- *(b+1)== 11 &&&&& *(b+i) ie *(b+2)== 6,

BEFORE *(b+1) = *(b+i)+5;==>> i=3 --- *(b+1)== 11 &&&&& *(b+i) ie *(b+3)== 8,
AFTER *(b+1) = *(b+i)+5; ==>> i=3 --- *(b+1)== 13 &&&&& *(b+i) ie *(b+3)== 8,

BEFORE *(b+1) = *(b+i)+5;==>> i=4 --- *(b+1)== 13 &&&&& *(b+i) ie *(b+4)== 10,

AFTER *(b+1) = *(b+i)+5; ==>> i=4 --- *(b+1)== 15 &&&&& *(b+i) ie *(b+4)== 10,
========================================================== OUTPUT:
2, 15, 6, 8, 10,
==========================================================

Only the value of *(b+1) ie b[1] ==4 is changing in memory till the end of the loop.

for(i=0;i<n;i++)
*(b+1)=*(b+i)+5;

Array={2,4,6,8,10};
*(b+1) indicates value at 0+1th index,that is value at array[1] becomes(4 becomes),b always be zero at all iteration,

i=0,In 1st iteration-2+5=7,
i=1,In 2 iteration-4+5=9,
i=2,In 3 iteration-6+5=11,
i=3,In 4 iteration-8+5=12,
i=4,In 5 iteration-10+5=15.

In these all iterations, only the value at array[1] changes at every iteration.

Change (a, 5); run and then the control is given back to the main function not stored. (i. E) it cannot update any value. So it prints the value of local variable.

Hence the output is 2, 4, 6, 8, 10.

*(b+1) = *(b+i)+5;
1. i = 0 => *(b+1) ie 4 is replaced by *(b+0)+5 ie 2+5
2. i = 1 => *(b+1) i.e 4 is replaced by *(b+1)+5 ie 4+5
3. i = 2 => *(b+1) i.e 9 is replaced by *(b+2)+5 ie 6+5
4. i = 3 => *(b+1) i.e 11 is replaced by *(b+3)+5 ie 8+5
5. i = 4 => *(b+1) i.e 15 is replaced by *(b+4)+5 ie 10+5

So, finally, 4 is replaced by 15 remaining are same.

Thank you @Pankaj Singh.

How the first element printed is 2?

The answer is B @Devendar.