Minimum of 10 numbers using for loop

Question

I have a simple question in this code:

#include <iostream>
using namespace std;
int main () {
int x, mn = 10000;
for (int i = 0 ; i<10 ; i++)
{
     cin>>x;
    if (x<mn)
        mn=x;

}
cout << mn;
return 0;
}

Why this outputs the minimum although if cases are true if we consider that the user won't enter a number greater than 10000?
My logic if the inputs are: in example 1, 2, 3, 4, 5, 6, 7, 8, 9, 10:

1<mn 
okay now mn =1; 
2<mn 
okay now mn =2; 
........ and so on, til mn = 10;

Why doesn't it output 10 (the last value)?
I will be thankful if you help.
PS: I also need a recommendation for better understanding how the code runs to me as a beginner.

I do not understand what you are asking. Does something in the provided code not work? If so, what isn't working? What do you expect and how does what you observe diffeR? — François Andrieux, Jan 12 '18 at 21:31
It does not output the last value because after the first iteration `mn` would be re-assigned from 10,000 to 1. After that all comparisons would produce a false. — Sergey Kalinichenko, Jan 12 '18 at 21:32
@FrançoisAndrieux No, the code runs but I need to understand how this outputs the minimum number? not the last value entered by the user. — Adam Hussein, Jan 12 '18 at 21:32
Debugger. Use a debugger. A debugger will allow you to single step through your program, watching values of variables. — Thomas Matthews, Jan 12 '18 at 21:34
See http://en.cppreference.com/w/cpp/language/if. This is a very basic question that may be better addressed by a [good book](https://stackoverflow.com/questions/388242/the-definitive-c-book-guide-and-list). — François Andrieux, Jan 12 '18 at 21:36

R Sahu · Answer 1 · 2018-01-12T21:51:27.327

3

Why doesn't it output 10 (the last value)?

Your understanding of the program is not correct.

You said:

1<mn 
okay now mn =1;

That is correct.

2<mn 
okay now mn =2;

That is not correct. At this time, value of mn is 1. Hence 2 < mn is not true. That would be true if you don't change of the value of mn but simply print x.

for (int i = 0 ; i<10 ; i++)
{
    cin>>x;
    if (x<mn)
      cout << x << endl;

}

edited Jan 12 '18 at 21:51

answered Jan 12 '18 at 21:36

R Sahu

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Thank you, that went so far to my mind, I don't expect that the value gonna change from 10000 to 1. That's what I don't understand. – Adam Hussein Jan 12 '18 at 21:38
@AdamHussein, the line `mn = x;` changes the value of `mn`. – R Sahu Jan 12 '18 at 21:39
Got it, he was not able to mentally trace execution the second time through when mn was already equal to 1. A debugger really helps with this. – Michael Dorgan Jan 12 '18 at 21:40

score 1 · Accepted Answer · edited Jan 12 '18 at 21:52

I have commented the code to help you understand each line.

#include <iostream>
using namespace std;
int main () {       // Start execution
int x, mn = 10000;  // Place a value of 10,000 into minimum - higher than the supposed maximum value that can be entered.  (This is not a good number to use)

for (int i = 0 ; i<10 ; i++)  // Loop 10 times
{
    cin>>x;    // Input a value into x
    if (x<mn)  // if x is less than mn (first loop it is 10,000) then...
        mn=x;  // Assign x into mn.

}
cout << mn;    // Output mn.  If no minimum was entered or found, this prints 10000
return 0;      // return 0
}

Minimum of 10 numbers using for loop

2 Answers2