Question

Assume m=5 and n=2 in each case. Answer the questions..

What are the output of the codes:

1) m*=n++; cout<<(--m)++<<","<<--n<<'\n';

2) Values of m and n after execution of codes: m*=n++;

3) What values of m, n will be after the statement: m+=--n; ?

4) What values of m and n will be after the statement: m*=n; ?

5) What values of m and n will be after the statement: m /=++n; ?

Answer 1

The output of the following programs are as follows:

1. 9,2

It is beacause initially the value of m = 5 and then m = m*n++

                                                                         -> m = 5*2++

                                                                                  = 10

        This is use then increment operator as ++ comes after the operand. Now, we execute the statement --m which means that decrement and then use. So, the current value of m is 10 which will be decremented to 9 due to the operator --m.

Regarding the second output after 9 that is 2. It comes because --n is in the pattern "--n", Now in the first statement that is

m*=n++, the value of n is 2 because it is post increment use and increment. Now, after this --n comes. As a result, the value of n will be decremented and then it will be suddenly decremented again due to pre-increment operator.

In other words, m*=n++;

                      The value of m is 10 and n is 2

                      cout<<(--m)++;   The value of m is 9 (pre increment ie, increment comes first).

                      Also, cout<<--n<<'\n';   The value of n is 2 (because now n is incremented due to n++ in the statement [ m*=n++] to 3 and again decremented to 2 [due to --n] in the statement.

2. values of m and n after execution of codes m*=n++ is

   m =   5*2   =   10

    because initial value of m is 5 and n will not increment and remain as 2 in this step .

The value of n is 2 itself.

3. The value of m and n after the statement m+=--n; is as follows:

    Let us say the current (initial value of m and n is 5 and 2 respectively)

     m = m+--n;

          = 5+(--2)

= 5 + 1

= 6 ,

The value of n is n=n-1=2-1=1

m=6 and n=1

because the value of n will be decremented ( decrement and use pre increment strategy).

4. The value of m and n after the statement m*=n; is as follows:

   Let us say the initial values of m and n are 5 and 2 respectively.

   m = m*n;

        = 5*2

       =   10

   n = 2

m = 10 and n =2

5. The values of m and n after statement m /=++n is as follows:

   Let us say the initial values of m=5 and n=2. Then,

   m = m/++n

       = 5/ (++2)

       = 5/ (2+1)     /* Pre Increment ie increment and then use */

        = 5/3

        = 1.6 , but

   if m is declared as int then the value of m is 1

if m is declared as float, then the value of m is 1.6

   However, the value of n is 3 ( n=n+1).

In general, questions from 1 to 5 is based on the concepts of post and pre increment/decrement operators.

The concept is this, in case of pre increment/decrement operator, initially the operand will be performed by the operation and the we will use it. As far as post increment/dcrement is concerned, the operand will be used first and then only increment or decrement will be done on the operand. You just understand these concepts.