[fariba_yoo]

syinamdar, on 01 February 2007 - 10:46 PM, said:

// Find maximum and minimum element in an arary.
#include<stdio.h>
int
main ()
{
  int i;
  int a[10] = { 10, 55, 9, 4, 234, 20, 30, 40, 22, 34 };
  int max = a[0];
  int min = a[0];

  for (i = 0; i < 10; i++)
	{
	  if (a[i] > max)
		{
		  max = a[i];
		}
	  else if (a[i] < min)
		{
		  min = a[i];
		}
	}
  printf ("Maximum element in an array : %d\n", max);
  printf ("Minimum element in an array : %d\n", min);

  return 0;
}

But how if I want to get numbers from user, like this:

int a[10];
int main()
{
	int Min=0;
	cout<<"Enter 10 number to find Minimum number: "<<endl;
	for (int i=0; i<10; i++)
	{
		cin>>a[i];
		if (Min>a[i])
		{
			Min=a[i];
		}
	}
	cout<<"Min is: "<<Min;
	getch();
}

here I have problem, I don't know how to initialize Min?

[KYA]

Code tags added and split from old topic.

As for your question, you could fill the array with data first then iterate through checking min's value against each index.

'min' is already initialized to zero when it is declared.

[tauit_dnmd]

fariba_yoo, on 12 July 2010 - 08:04 AM, said:

syinamdar, on 01 February 2007 - 10:46 PM, said:

// Find maximum and minimum element in an arary.
#include<stdio.h>
int
main ()
{
  int i;
  int a[10] = { 10, 55, 9, 4, 234, 20, 30, 40, 22, 34 };
  int max = a[0];
  int min = a[0];

  for (i = 0; i < 10; i++)
	{
	  if (a[i] > max)
		{
		  max = a[i];
		}
	  else if (a[i] < min)
		{
		  min = a[i];
		}
	}
  printf ("Maximum element in an array : %d\n", max);
  printf ("Minimum element in an array : %d\n", min);

  return 0;
}

But how if I want to get numbers from user, like this:

int a[10];
int main()
{
	int Min=0;
	cout<<"Enter 10 number to find Minimum number: "<<endl;
	for (int i=0; i<10; i++)
	{
		cin>>a[i];
		if (Min>a[i])
		{
			Min=a[i];
		}
	}
	cout<<"Min is: "<<Min;
	getch();
}

here I have problem, I don't know how to initialize Min?

Because you init for Min=0, so if you input a number (>0) ,it always return 0.
To solve you can you a variable BOOL to check and init value for Min
Such as: bool first;
Here.

#include<iostream>
using namespace std;

int a[10];
int main()
{
	bool first=true; // input first time.
	int Min=0;
	cout<<"Enter 10 number to find Minimum number: "<<endl;
	for (int i=0; i<10; i++)
	{
		cin>>a[i];
		if(first)
		{
			first=false; 
			Min=a[i]; //init for Min in the first time .
		}
		else if(Min>a[i]) Min=a[i];
	}
	cout<<"Min is: "<<Min;
	return 0; //you must return a value .
}

This post has been edited by tauit_dnmd: 12 July 2010 - 09:25 AM

[CTphpnwb]

No need to use a boolean:

#include<iostream>
using namespace std;

int a[10];
int main()
{
	int Min=0;
	cout<<"Enter 10 number to find Minimum number: "<<endl;
	for (int i=0; i<10; i++)
	{
		cin>>a[i];
		if( !i || Min>a[i] )
		{
			Min=a[i];
		}
	}
	cout<<"Min is: "<<Min;
	return 0; //you must return a value .
}

This post has been edited by CTphpnwb: 12 July 2010 - 09:51 AM

[fariba_yoo]

CTphpnwb, on 12 July 2010 - 08:50 AM, said:

No need to use a boolean:

#include<iostream>
using namespace std;

int a[10];
int main()
{
	int Min=0;
	cout<<"Enter 10 number to find Minimum number: "<<endl;
	for (int i=0; i<10; i++)
	{
		cin>>a[i];
		if( !i || Min>a[i] )
		{
			Min=a[i];
		}
	}
	cout<<"Min is: "<<Min;
	return 0; //you must return a value .
}

Thanks a lot, it worked, can you please explain about this part: if( !i || Min>a[i] )
I couldn't understand !i part....

[KYA]

'i' is an integer. '!' is the logical NOT

0 and 1 can be equated as false/true

Thus NOT zero = true and NOT [any other number] = false

[OnlineCop]

Can you use negative numbers?

If negatives are NOT allowed, it may be easier to do this:

int main()
{
  unsigned int Min = (unsigned int) -1; // Basically, a VERY LARGE positive number

  ... // all the rest of your code
}

What this does is that the user will enter a number. Obviously, it is smaller than "the largest possible integer value" (which the (unsigned int) -1 is set to), so therefore, the "new" value becomes the new Min.

Now, if negatives ARE valid, you could set it to the largest positive number available (in this case, you can change it to the hex value 0x7FFFFFFF). It will still be greater than whatever value they enter, and therefore, you're still in the clear.

[fariba_yoo]

Thanks, it was useful