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5.1 The compound condition y%4 == 0 && (y % 100 != 0 || 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. Convert the reference parameter into a return value. For example, the function void f(int& n) { n *= 2; } is equivalent to the function int g(int n) { return 2*n; } The two functions are invoked differently: int x=22, y=44; f(x); // double the value of x y = g(y); // double the value of y But in both cases, the effect is to double the value of the parameter. This is similar to Example 5.2: int main() { for (float x=0; x < 1; x += 0.1) cout << cos(2*x) << '\t' << 2*cos(x)*cos(x) - 1 << endl; } 1 1 0.980067 0.980067 0.921061 0.921061 0.825336 0.825336 0.696707 0.696707 0.540302 0.540302 0.362358 0.362358 0.169967 0.169967 -0.0291997 -0.0291997 -0.227202 -0.227202 The equal values show that the identity is true for the 10 values of x tested. This is similar to Example 5.2: int main() { for (double x=0; x < 2; x += 0.2) { double s=sin(x); double c=cos(x); cout << s*s << "\t" << c*c << "\t" << s*s+c*c << endl; } }
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0 1 1 0.0394695 0.96053 1 0.151647 0.848353 1 0.318821 0.681179 1 0.5146 0.4854 1 0.708073 0.291927 1 0.868697 0.131303 1 0.971111 0.0288888 1 0.999147 0.000852612 1 0.948379 0.0516208 1 0.826822 0.173178 1 This is similar to Example 5.2: int main() { double b=2; double lg2=log(2); for (double x=0; x < 2; x += 0.2) cout << pow(b,x) << "\t" << exp(x*lg2) << endl; } 1 1 1.1487 1.1487 1.31951 1.31951 1.51572 1.51572 1.7411 1.7411 2 2 2.2974 2.2974 2.63902 2.63902 3.03143 3.03143 3.4822 3.4822 4 4 This tests a function that returns the minimum of four integers: int min(int,int,int,int); int main() { cout << "Enter four integers: "; int w, x, y, z; cin >> w >> x >> y >> z; cout << "Their minimum is " << min(w,x,y,z) << endl; } int min(int n1, int n2, int n3, int n4) { int min=n1; if (n2 < min) min = n2; if (n3 < min) min = n3; if (n4 < min) min = n4; return min; } Enter four integers: 44 88 22 66 Their minimum is 22 This tests a function that returns the maximum of three integers: int max(int,int,int); int main() { cout << "Enter three integers: "; int x, y, z; cin >> x >> y >> z; cout << "Their maximum is " << max(x,y,z) << endl; }
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int max(int, int); int max(int x, int y, int z) { int max(int,int); return max(max(x,y),z); } int max(int x, int y) { // returns the maximum of the two given integers: if (x < y) return y; else return x; } Enter three integers: 44 88 66 Their maximum is 88 This tests a function that returns the minimum of four integers: int min(int,int,int,int); int main() { cout << "Enter four integers: "; int w, x, y, z; cin >> w >> x >> y >> z; cout << "Their minimum is " << min(w,x,y,z) << endl; } int min(int,int); int min(int n1, int n2, int n3, int n4) { int m12=min(n1,n2); int m34=min(n3,n4); return (m12 < m34 m12 : m34); } int min(int m, int n) { return (m < n m : n); } Enter four integers: 44 88 22 66 Their minimum is 22 This tests a function that returns the average of four numbers: double ave(double,double,double,double); double main() { cout << "Enter four numbers: "; double w, x, y, z; cin >> w >> x >> y >> z; cout << "Their average is " << ave(w,x,y,z) << endl; } double ave(double x1, double x2, double x3, double x4) { return (x1 + x2 + x3 + x4)/4.0; } Enter four numbers: 44 88 22 66 Their average is 55 This tests a function that returns the average of four or fewer numbers: double ave(double,double=0,double=0,double=0); double main() { cout << "Enter four non-zero numbers: "; double w, x, y, z; cin >> w >> x >> y >> z; cout << "The average of the first one is " << ave(w) << endl;
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