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evaluate the constant polynomial a03, or by p ( x , aO , al ) to evaluate the first-degree polynomial a0 + alx,orby p(x, aO, al, a2 ) to evaluate the second-degree polynomial a0 + a+ + a2x2. Note how the default values are given in the function prototype. Here is the output from the test run:
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For example the call p (x, 7 , 6 , 5 ) , which is equivalent to the call p ( the second degree polynomial 7 + 6 x + 5 x2.
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7 , 6 , 5 , 0 ) , evaluates
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In the example above, the function may be called with 2,3,4, or 5 arguments. So the effect of allowing default parameter values is really to allow a variable number of actual parameters passed to the function. If a function has default parameter values, then the function s parameter list must show all the parameters with defau t values to the right of all the parameters that have no default values, like this:
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void f(int a, void g(int a, int b, int c=4, int d=7, int e=3); int d, int e=3); // OK // ERROR int b=2, int c=4,
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The optional parameters must all be listed last.
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4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 4.9 What are the advantages of using functions to modularize a program What is the difference between a function s declaration and its definition Where can the declaration of a function be placed When does a function need an include directive What is the advantage of putting a function s definition in a separate file What is the advantage of compiling a function separately What are the differences between passing a parameter by value and by reference What are the differences between passing a parameter by reference and by constant reference Why is a parameter that is passed by value referred to as read-only Why is a parameter that is passed by reference referred to as read-write
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4.10 What is wrong with the following declaration:
int f(int a, int b=O, int c);
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FUNCTIONS
Solved Problems
4.11 In Example 4.13, the following expression was used to test whether y is a leap year: Y % 4 == 0 &SC y % 100 != 0 I I y % 400 == 0 This expression is not the most efficient form. If y is not divisible by 4, it will still test the cond i t i o n y % 40 0 == 0 which would have to be false. C++ implements short circuiting, which means that subsequent parts of a compound condition are tested only when necessary. Find an equivalent compound condition that is more efficient due to short circuiting.
T h e c o m p o u n d condition
y%4 == 0 &SC (y % 100 != 0 11 y % 400 == 0)
is equivalent and more efficient. The two can be seen to be equivalent by checking their values in the four possibilities, represented by the four y values 1995, 1996, 1900, and 2000. This condition is more efficient because if y is not divisible by 4 (the most likely case), then it will not test y further. 4.12 Describe how a void function with one reference parameter can be converted into an equiva-
lent non-void function with one value parameter.
Convert the reference parameter into a return value. For example, the function
void f(int& n) -t n *, 20I
is equivalent to the function
int g(int n) -t return 2*n;
These two functions are invoked differently:
int x = 22, y = 33; f (x> ; Y = g(y);
But in both cases, the effect is to double the value of the parameter.
Solved Programming Problems
4.13 Write a simple program like the one in Example 4.2 to check the identity: = - 1.
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This is similar to Example 4.2: #include #include main0 for (float x = 0; x < 1; x += 0.1) tout -CC cos(2*x) << '\t' << 2*cos(x)*cos(x) <iostream.h> <math.h>
- 1 << endl;
Q.98006 T 0*92106~ fL825336 0.696707 0.5~0302 0,362X58 O*Z6.9967 -0.0291997 -0 I227202
U,9aoo67 0*92106l * 0.825336 -0 a 696707 0.540302 0,362358 0 JAi9.96 7 -0A2915397 -0.227202
Each value in the first column matches the corresponding value in the second column, showing that the identity is true for the 10 values of x tested. 4.14 A more efficient way to compute the permutations function P(n,k) is by the formula
P(n,k) = (n) (n-l) (n-2)...(n-k+2) (n-k+l)
This means the product of the k integers from n down to n - k + 1. Use this formula to rewrite and test the perm ( > function from Example 4.9.
To compute a product of k integers, we use a for loop that iterates k times. Each time, p is multiplied by n which is then decremented. The result is that 1 is multiplied by n, n- 1, n- 2, etc., downto n-k+l: int perm(int, int);
main0 { for (int i = -1; i < 8; i++) { for (int j = -1; j <= i+l; j++) tout << " H << perm(i,j); tout << endl; >
// Returns P(n,k), the number of permutations of k from n: int perm(int n, int k)
if (n < 0 II k < 0 II k > n) return 0; int p = 1; for (int i = 1; i <= k; i++, n-4 p *-n; return p; 1
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