program using array need to find the position of the number

Sep 28, 2012 at 4:27pm
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
Sep 28, 2012 at 6:22pm
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#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;
}


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 Sep 28, 2012 at 6:24pm
Sep 30, 2012 at 2:05am
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!
Sep 30, 2012 at 3:11am
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//from line 16
cin >> innum;
for(int i=0; i<n; i++)
{
   if(a[i] == innum)
   {
       cout << "Number " << innum << " is present at position " << i;
   }
}

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