Type Conversion in Assignments

It may so happen that the type of the expression and the type of the variable on the left-hand side of the assignment operator may not be same. In such a case the value of the expression is promoted ordemoted depending on the type of the variable on left-hand side of =.
For example, consider the following assignment statements.

int i ;
float b ;
i = 3.5 ;
b = 30 ;

Here in the first assignment statement though the expression’s value is a float (3.5) it cannot be stored in i since it is an int. In such a case the float is demoted to an int and then its value is stored. Hence what gets stored in i is 3. Exactly opposite happens in the next statement. Here, 30 is promoted to 30.000000 and then stored in b, since b being a float variable cannot hold anything except a float value.
Instead of a simple expression used in the above examples if a complex expression occurs, still the same rules apply. For example, consider the following program fragment.

float a, b, c ;
int s ;
s = a * b * c / 100 + 32 / 4 - 3 * 1.1 ;

Here, in the assignment statement some operands are ints whereas others are floats. As we know, during evaluation of the expression the ints would be promoted to floats and the result of the expression would be a float. But when this float value is assigned to s it is again demoted to an int and then stored in s.
Observe the results of the arithmetic statements shown in Figure 1.7. It has been assumed that k is an integer variable and a is a real variable.

Note that though the following statements give the same result, 0, the results are obtained differently.
k = 2 / 9 ;
k = 2.0 / 9 ;
In the first statement, since both 2 and 9 are integers, the result is an integer, i.e. 0. This 0 is then assigned to k. In the second statement 9 is promoted to 9.0 and then the division is performed. Division yields 0.222222. However, this cannot be stored in k, k being an int. Hence it gets demoted to 0 and then stored in k.