### program using array need to find the position of the number

please take a look at my current work. thanks

#include <iostream>
using namespace std;

int main()
{
int position, innum, n, i, a[100];
cout << "Please input a number: ";
cin >> n;
cout << "\nPlease input the " << n << " numbers: \n";
for (i=0; i<n; i++)
cin >> a[i];
cout << "\nThe following are the list of numbers: \n";
for (i=0; i<n; i++)
cout << " " << a[i];
cout << "\n\nEnter number to be search: \n";
cin >> innum;
innum == a[i];//nevermind this much for im still in deep confusion as to how to go about this
cout << "Number " << innum << "is present at position " << innum;

system("pause>0");
return 0;
}

this should be the output:
input n: 8 (n) 8 is a variable

enter 8 numbers
100 53 84 320 3 12 43 121

enter number to be search: 320
number 320 is present at position 4
number 320 is present at position 8
Code tags...
 ``12345678910111213141516171819202122`` ``````#include using namespace std; int main() { int position, innum, n, i, a[100]; cout << "Please input a number: "; cin >> n; cout << "\nPlease input the " << n << " numbers: \n"; for (i=0; i> a[i]; cout << "\nThe following are the list of numbers: \n"; for (i=0; i> innum; innum == a[i];//nevermind this much for im still in deep confusion as to how to go about this cout << "Number " << innum << "is present at position " << innum; system("pause>0"); return 0; }``````

 innum == a[i];//nevermind this much for im still in deep confusion as to how to go about this

Just loop over the a[] array and over each iteration test if a[i] == innum.
If it is == then `cout << "Number " << innum << "is present at position " << i;`

You should also always (if possible) have 'for' loops like this:
`for (int i=0; i<n; i++) //declare 'int i' in the loop itself to keep things local `
Last edited on
do you mean im gonna have to put an if else loop before or after the for loop?
please bear with me with this one thanks!
 ``12345678910111213`` ``````//from line 16 cin >> innum; for(int i=0; i0"); return 0; }``````
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