Function Composition

Section 4.12 Function Composition

You have probably learned how to evaluate simple expressions like \(\sin(\pi/2)\) and \(\log(1/x)\text{.}\) First, you evaluate the expression in parentheses, which is the argument of the function. Then you can evaluate the function itself, either by hand or by punching it into a calculator.

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This process can be applied repeatedly to evaluate more-complex expressions like \(\log(1/\sin(\pi/2))\text{.}\) First we evaluate the argument of the innermost function (\(\pi/2 = 1.57\)), then evaluate the function itself (\(\sin(1.57) = 1.0\)), and so on.

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Just as with mathematical functions, C++ functions can be composed to solve complex problems. That means you can use one function as part of another. In fact, you can use any expression as an argument to a function, as long as the resulting value has the correct type:

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const double PI = acos(-1.0);  //calculate an accurate value of PI
double x = cos(angle + PI / 2.0);

This statement divides PI by 2.0, adds the result to angle, and computes the cosine of the sum.

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Here is another example you may remember from geometry. Given the lengths of legs a and b in a right triangle, you can find the length of the hypotenuse c with \(\sqrt{a^2 + b^2}\text{.}\) Here is that translated to C++

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double sideA = 3;
double sideB = 4;
double sideC = sqrt( pow(sideA, 2) + pow(sideB, 2) );

Before sqrt can do its job, we need to compute its argument by doing the +. But before that can happen, the two pow calls must run. So, they happen first. Then their results are added together. Finally, the square root is taken.

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Warning 4.12.1.

Just because you can use composition to nest a large number of steps into one line of code, does not mean that you should. This is perfectly legal:

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sqrt(pow(hypot(x2 - x1, y2 - y1), 2) + pow(hypot(x4 - x3, y4 - y3), 2) - 2 * hypot(x2 - x1, y2 - y1) * hypot(x4 - x3, y4 - y3) * cos(angleC));

But it is hard to read and will be even harder to debug if you make a mistake somewhere on the line. It also redoes the same work in multiple spots. A better approach would be to write the code as a series of steps that calculate the desired value.

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double sideA = hypot(x2 - x1, y2 - y1);
double sideB = hypot(x4 - x3, y4 - y3);
...

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Checkpoint 4.12.1.

cmath Which expression correctly computes \(\log_2{(10^8)}\text{?}\) Here is the cmath library for reference.

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log2(pow(8, 10))
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That computes 8 to the 10th power.
log2(pow(10, 8))
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log(2, pow(8, 10))
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There is no log function that takes two parameters.
log(2, pow(10, 8))
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There is no log function that takes two parameters.
pow(log2(10), 8)
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That first does \(\log{2}{10}\text{.}\) We want to first do the pow.

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