array item assigning in sorting functions of c++

Question

its not correctly assigning items in an array during the sorting function of selection sort in c++.what is the correct way?

int a[] = { 22, 91, 35, 78, 10, 8, 75, 99, 1, 67 };
int arr_len = sizeof(a)/sizeof(a[0]);

for(int index = 0; index < arr_len - 1; index++)
{
    for(int n = index + 1; n < arr_len; n++)
    {
        if(a[index] > a[n])
        {
            a[index] = a[n];
            a[n] = a[index];
        }
    }
}

the result of my program gave me like this: Array in sorted order: 1 1 1 1 1 1 1 1 1 67

Two swap two elements use a "temp" value. temp=a; a=b; b= temp; — Ripi2
– Ripi2, Commented Sep 6, 2019 at 20:01
Unrelated: If t you're compiling to a recent standard, std::size can be used to replace sizeof(a)/sizeof(a[0]) — user4581301
– user4581301, Commented Sep 6, 2019 at 20:57

Remy Lebeau · Accepted Answer · 2019-09-06 22:33:22Z

1

Let's say that you want to swap a and b. Let's say that the values are:

a = x
b = y

The first step of your attempt was to assign a = b. After such operation, the situation would be:

a = y
b = y

How could you at this point assign x to b? Think about it.

You can't. The original value of a has been lost. So, clearly this first step leads to a dead end.

The solution: Introduce a new variable, assign a onto that variable, assign b onto a, and finally assign the new variable onto b.

temp = a
a = b
b = temp

edited Sep 6, 2019 at 22:33

Remy Lebeau

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answered Sep 6, 2019 at 20:05

eerorika

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