max - naive tool for printing maximum floating-point value from given set

README.md

Example usage:

~ $ g++ max.cc -o max
~ $ max 10 20 0 0 2 3 1 # reads input from command line
20
~ $ (echo 10; echo 20; echo 0; echo 0; echo 2) | max -i # or from stdin
20
~ $ max -0.3 -0.5 -0.1 -2.3
-0.1000
~ $ solve_linear 2 11
-5.5000

max.cc

#include <iostream>
#include <cstdlib>
#include <iomanip>
#include <limits>
#include <vector>
#include <string>

using namespace std;

void die_usage() {
	cerr << "usage: max [-i] [integer...]" << endl;
	exit(1);
}

int main(int argc, char *argv[]) {
	if (argc == 1) {
		die_usage();
	}
	// Read input.
	string line;
	vector<string> lines;
	if (string(argv[1]) == "-i") {
		while (getline(cin, line)) {
			lines.push_back(line);
		}
	} else {
		lines = vector<string>(argv+1, argv+argc);
	}
	// Convert to ints.
	vector<double> nums;
	for (size_t i = 0; i < lines.size(); ++i) {
		nums.push_back(0.0);
		if (sscanf(lines[i].c_str(), "%lf", &nums[i]) != 1)  {
			cerr << lines[i] << " is not a valid integer" << endl << endl;
			die_usage();
		}
	}
	// Count max.
	double max = -numeric_limits<double>::max();
	for (size_t i = 0; i < nums.size(); i++) {
		if (nums[i] > max) {
			max = nums[i];
		}
	}
	cout << fixed << setprecision(4) << max << endl;
	return 0;
}

solve_linear.cc

#include <iostream>
#include <cstdlib>
#include <iomanip>
#include <limits>
#include <vector>
#include <string>
#include <cmath>

using namespace std;

void die_usage() {
	cerr << "usage: solve_linear [a] [b]" << endl;
	exit(1);
}

bool equals(double i, double j) {
	    return fabs(i - j) < numeric_limits<double>::epsilon();
}

int main(int argc, char *argv[]) {
	if (argc != 3) {
		die_usage();
	}
	// Read input.
	const vector<string> args = vector<string>(argv+1, argv+argc);
	// Convert to ints.
	vector<double> nums;
	for (size_t i = 0; i < args.size(); ++i) {
		nums.push_back(0.0);
		if (sscanf(args[i].c_str(), "%lf", &nums[i]) != 1)  {
			cerr << args[i] << " is not a valid double value" << endl << endl;
			die_usage();
		}
	}
	// Solve linear and print.
	if (!equals(nums[0], 0.0)) {
		cout << fixed << setprecision(4) << (-nums[1]/nums[0])  << endl;
	} else if (equals(nums[1], 0.0)) {
		cout << "solve_linear: too many solutions for a=" << args[0] << ", b=" << args[1] << endl;
	} else {
		cout << "solve_linear: no solution for a=" << args[0] << ", b=" << args[1] << endl;
	}
	return 0;
}