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The next statement z = x + y; evaluates
the expression on right hand side. It takes values stored in variables
x and y (which are 5 and 10 respectively), adds them and by using the
assignment operator (=), puts the value of the result, which is 15 in
this case, to the memory location labeled as z.
Here a thing to be noted is that
the values of x and y remains the same after this operation. In arithmetic
operations the values of variables used in expression on the right hand
side are not affected. They remain the same. But a statement like x
= x + 1; is an exceptional case. In this case the value of x is changed. The next line cout << “ x
= “ ; is simple. It just displays ‘ x = ‘ on the screen.
Now we want to display the value
of x after ‘x =’. For this we write the statement cout << x ; Here comes the affect of data type
on cout. The previous statement cout << “x = “ ; has a character
string after << sign and cout simply displays the string. In the
statement cout << x; there is a variable name x. Now cout will
not display ‘x’ but the value of x. The cout interprets that x is a
variable of integer type, it goes to the location x in the memory and
takes its value and displays it in integer form, on the screen. The
next line cout << ”y =”; displays ‘ y = ‘ on the screen. And line
cout << y; displays the value of y on the screen. Thus we see
that when we write something in quotation marks it is displayed as it
is but when we use a variable name it displays the value of the variable
not name of the variable. The next two lines cout << “z = x +
y = ”; and cout << z; are written to display ‘z = x + y = ’ and
the value of z that is 15. Now when we execute the program
after compiling, we get the following output
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