C++ array operations - c++

Question

I want from the program to add 3 to elements which are greater than 3 and print them. It takes so much time that I couldn't see the result. Also, when I change n to 8 in loops directly, it gives a result; however, it's not related with what I want. How can I correct this code and improve that?
#include <iostream>
using namespace std;
int main()
{
int a[8] = {-5, 7, 1, 0, 3, 0, 5, -10};
int b[8];
int n= sizeof(a);
for( int i=1; i<=n; i++) {
if (a[i]>3) {
b[i] = a[i] + 3;
}
else {
b[i]= a[i];
}
}
for (int i = 1; i <= n; i++)
cout << b[i];
return 0;
}

Answer 1

The sizeof() function (int n = sizeof(a)) gives 32 because array 'a' contains 8 elements & each element is of 'int' type whose size is 4 byte in memory thats why it returns 32 in 'n' variable.so you must divide the value of 'n' with the size of integer.
Secondly the index of array starts with the zero '0' to one less than the length of array not with the 1 to length of array .
Try the below code ! I am also attach the output of the code .
#include <stdio.h>
#include <iostream>
using namespace std;
int main()
{
int a[8] = { -5, 7, 1, 0, 3, 0, 5, -10 };
int b[8];
int n = sizeof(a)/sizeof(int);
for (int i = 0; i < n; i++) {
if (a[i]>3) {
b[i] = a[i] + 3;
}
else {
b[i] = a[i];
}
}
for (int i = 0; i < n; i++)
cout << b[i]<<endl;
return 0;
}

Answer 2

The statement in your program int n=size(a) returns the total bytes occupied in memory for a. i.e int occupies 4 bytes and a is an array contains 8 elements so 8X4 = 32 .but while accessing the array elements using loop you are specifying i<=n meains i<=32 but there is only 8 elements but you are trying to access 32 elements which indicates that you are trying to access the elements more than 8.
exeutes the following code
#include <iostream>
using namespace std;
int main()
{
int a[8] = {-5, 7, 1, 0, 3, 0, 5, -10};
int b[8];
int n=sizeof(a);
cout<<"\n Value of n is : "<<n;
return 0;
}
Output
Value of n is : 32
if you specify the exact number of array size your program will work properly.
#include <iostream>
using namespace std;
int main()
{
int a[8] = {-5, 7, 1, 0, 6, 0, 8, -10};
int b[8];
for( int i=1; i<8; i++)
{
if (a[i]>3)
{
b[i] = a[i] + 3;
}
else
{
b[i]= a[i];
}
}
cout<<"\n Values in a array";
cout<<"\n -----------------\n";
for (int i = 1; i <8; i++)
cout << "\t"<<a[i];
cout<<"\n Values in b array";
cout<<"\n -----------------\n";
for (int i = 1; i <8; i++)
cout << "\t"<<b[i];
return 0;
}
OUTPUT
Values in a array
-----------------
7 1 0 6 0 8 -10
Values in b array
---------------
10 1 0 9 0 11 -10
I hope that you understand the concept.Thank you

Answer 3

Your int n = sizeof(a); doesn't works as you intend.
I think you want to get the size of array (i.e. 8).
But you gets the size in bytes of elements (e.g. 32, integer size or could be different depending of your system's architecture).
Change to int n = 8, will solve your problem.
Also note that for( int i=1; i<=n; i++) will get an "out of array" element.

Answer 4

Sizeof function gives the value of the total bits of memory the variable occupy
Here in array it stores 8 integer value that has 2bit size for each thus it returns 32