Get max value

[Alauda]

Hi,

I have a line in my code which calculates the sum of all a's:

a_total += a

I would like to adapt the line such that I obtain the maximum, instead of the sum. So something like:

a_max max= a

Of course this does not work like this. What would be the smartest way to do this? Thank you!

Cheers, Chris

[Chervil]

I don't know whether this is particularly smart...
a_max = (a>a_max) ? a : a_max;

[Bourgond Aries]

Since you have multiple "a"s, let's assume it to be an array.

int largest = a[0];
for (int i = 1; i < a_size; i++)
{
    if (a[i] > largest)
        largest = a[i];
}

Now you have iterated overy every element of a, and you've found the largest value and stored it in largest.

[Alauda]

Hi!

Thank you both very much! The first answer from Chervil compiles properly, but I suspect it to be wrong, since a_max returns a lower value compared to when calculating the average of all a's.

The second answer (Bourgond) looks like it should work, but I cannot compile it. I imagine this occurs because in my code "a" is declared as "double". Simply exchanging "int" with "double" does not work either, I get:

error: invalid types ‘double[int]’ for array subscript

I'll keep trying (absolute beginner). Thanks in any case!

Cheers, Chris

[Smac89]

#include <iostream>
using namespace std;

int main()
{
   int max=0;
   for (auto& k: {9,5,4,3,90,-4,91,23,29,-39})
      max = (k > max ? k : max);
   
   cout << max << endl;
   return 0;
}

compile with -std=c++11

[Bourgond Aries]

Maybe you could post your code? Seems like there's some confusion regarding 'a'.

[vlad from moscow]

If I have correctly understood you you need to find the maximum value among enetered numbers.

This is being done in two steps. At first you assign the first entered number to a_max.

a_max = a;

Then when a next number is entered you should compare it with the current value of a_max:

if ( a_max < a ) a_max = a;

[Alauda]

Hi everyone,

Sorry for the silence, I got distracted. Yes, Vlad, your approach works! Thanks to everyone!

All the best,
Chris