"Compose two functions, and you still have a function. Compose ten, or a hundred, and it is still a function mapping values to values."
The problem being that when you compose two functions the number of possible input combinations to function 1 multiplied by the number of possible input combinations to function 2.
When you have 10 functions, each with two inputs, combining them gives you 2^10 combination to test - 1024 combinations.
Whereas if you test each one separately you can test 20 combinations, and get the same coverage.
I don’t know what you’re talking about, but it isn’t function composition. Function composition consumes outputs as it produces inputs, in equal measure — in f(g(x)), the domain of f is bounded by the codomain of g, which is itself bounded by x. If g has three branches, but f always outputs a value that takes the first branch, then g will only ever follow one branch in your test; you won’t get a combinatoric explosion of possible branches taken.
You might be thinking more of testing several expressions independently, e.g. assert([f(x), g(x)] == […]). This does what you’re talking about, but I don’t think anybody (including the author of the post) was recommending this.
Comments
"Compose two functions, and you still have a function. Compose ten, or a hundred, and it is still a function mapping values to values."
The problem being that when you compose two functions the number of possible input combinations to function 1 multiplied by the number of possible input combinations to function 2.
When you have 10 functions, each with two inputs, combining them gives you 2^10 combination to test - 1024 combinations.
Whereas if you test each one separately you can test 20 combinations, and get the same coverage.
I don’t know what you’re talking about, but it isn’t function composition. Function composition consumes outputs as it produces inputs, in equal measure — in f(g(x)), the domain of f is bounded by the codomain of g, which is itself bounded by x. If g has three branches, but f always outputs a value that takes the first branch, then g will only ever follow one branch in your test; you won’t get a combinatoric explosion of possible branches taken.
You might be thinking more of testing several expressions independently, e.g. assert([f(x), g(x)] == […]). This does what you’re talking about, but I don’t think anybody (including the author of the post) was recommending this.