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sortix--sortix/utils/calc.cpp

398 lines
10 KiB
C++

/******************************************************************************
COPYRIGHT(C) STEVEN DOUGHERTY 2011.
COPYRIGHT(C) JONAS 'SORTIE' TERMANSEN 2011.
This program is free software: you can redistribute it and/or modify it
under the terms of the GNU General Public License as published by the Free
Software Foundation, either version 3 of the License, or (at your option)
any later version.
This program is distributed in the hope that it will be useful, but WITHOUT
ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or
FITNESS FOR A PARTICULAR PURPOSE. See the GNU General Public License for
more details.
You should have received a copy of the GNU General Public License along
with this program. If not, see <http://www.gnu.org/licenses/>.
calc.cpp
A simple reverse polish calculator.
******************************************************************************/
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <errno.h>
//Stack, pending standard library implementation. All in one file due to linking
// annoyances with the makefile.
template <typename T>
class Stack {
public:
//Is its own iterator.
/* MODIFIES: this.
* EFFECTS: Self-iterator: sets to first element and returns it.
*/
const T *begin() {
currentIter = head;
return next();
}
/* MODIFIES: this.
* EFFECTS: Self-iterator: returns element from the list and advances.
* If all elements have been iterated, returns NULL. Does not
* have invalidation - be careful!
*/
const T *next() {
if (!currentIter) return NULL;
const T *element = currentIter->o;
currentIter = currentIter->next;
return element;
}
/* EFFECTS: Returns true if stack is empty, false otherwise.
*/
bool isEmpty() const {
return !head;
}
/* MODIFIES: this.
* EFFECTS: Pushes o onto the top of the stack.
*/
void push(T *o) {
head = new node(head, o);
}
/* MODIFIES this
* EFFECTS Pops and returns the top element. Returns NULL if empty.
*/
T *pop() {
if (isEmpty()) return NULL;
node *oldHead = head;
T *element = oldHead->o;
head = oldHead->next;
delete oldHead;
return element;
}
Stack() : head(NULL) , currentIter(NULL) {}
Stack(const Stack &s) : head(NULL) , currentIter(NULL) {
s.begin();
copyAll(s.head);
}
Stack &operator=(const Stack &s) {
if (this == &s) return;
removeAll();
copyAll(s.head);
}
~Stack() {
removeAll();
}
private:
struct node {
node(node *next, T *o) : next(next), o(o) {}
node *next;
T *o;
};
node *head;
node *currentIter;
/* MODIFIES: this.
* EFFECT: called by copy constructor and operator= to copy elements
* following from the given node into the local instance.
*/
void copyAll(const node *n) {
if (n) {
copyAll(n->next);
push(new T(*(n->o)));
}
}
void removeAll() {
while (!isEmpty()) delete pop();
}
};
/* MODIFIES: stack.
* EFFECTS: Pushes value onto the stack.
*/
void push(long *value, Stack<long> &stack) {
stack.push(value);
}
/* MODIFIES: stack.
* EFFECTS: Pushes value onto the stack.
*/
void push(long value, Stack<long> &stack) {
stack.push(new long(value));
}
/* MODIFIES: operand, stack, standard output.
* EFFECTS: Attempts to set operand to the top value on the stack. If there
* are no operands on the stack, prints an error message and aborts
* the process with exit code 1.
*/
void pop(long *&operand, Stack<long> &stack) {
if (stack.isEmpty()) {
printf("%s: not enough operands\n", program_invocation_name);
exit(1);
}
operand = stack.pop();
}
/* MODIFIES: operand1, operand2, stack, standard output.
* EFFECTS: Attempts to remove two operands from the top of the stack, setting
* operand1 to the first one removed and operand2 to the second. If
* there are fewer than two enough operands on the stack, prints
* an error message and aborts the process with exit code 1.
*/
void popTwo(long *&operand1, long *&operand2, Stack<long> &stack) {
pop(operand1, stack);
pop(operand2, stack);
}
/* MODIFIES: stack, standard output.
* EFFECTS: Attempts to replace the top two operands on the stack with the
* result returned by func(first, second).
*/
//Lambda functions would be wonderful for calling this function.
//http://www2.research.att.com/~bs/C++0xFAQ.html#lambda
void apply(long(*func)(long, long), Stack<long> &stack) {
long *operand1, *operand2;
popTwo(operand1, operand2, stack);
push(func(*operand1, *operand2), stack);
delete operand1;
delete operand2;
}
/* EFFECTS: Returns first + second.
*/
long add(long first, long second) { return first + second; }
/* EFFECTS: Returns subtracting "the first number from the second" (pg. 5)
*/
long subtract(long first, long second) { return second - first; }
/* EFFECTS: Returns first * second.
*/
long multiply(long first, long second) { return first * second; }
/* MODIFIES: stack, standard output.
* EFFECTS: Attempts to replace the top two operands on the stack with the
* result of dividing the second one by the first.
* If the first number is zero, prints an division by zero error and
* exits the process with exit code 1.
*/
void divide(Stack<long> &stack) {
/* Can't use apply for this because it requires a check for the
* denominator being zero. It could have an error-checking function
* pointer, but that would muddy things as no other operators have such
* conditions.
*/
long *first, *second;
popTwo(first, second, stack);
if (*first == 0) {
printf("%s: division by zero\n", program_invocation_name);
exit(1);
}
push(*second / *first, stack);
delete first;
delete second;
}
/* MODIFIES: stack, standard output.
* EFFECTS: Attempts to negate the top element on the stack.
*/
void negate(Stack<long> &stack) {
long *operand;
pop(operand, stack);
push(-(*operand), stack);
delete operand;
}
/* MODIFIES: stack, standard output.
* EFFECTS: Attempts to push a copy of the top element on top of the already
* existing one.
*/
void duplicate(Stack<long> &stack) {
long *operand;
pop(operand, stack);
//Dereferencing creates new instance.
push(*operand, stack);
push(operand, stack);
}
/* MODIFIES: stack, standard output.
* EFFECTS: Attempts to reverse the first two elements on the stack.
*/
void reverse(Stack<long> &stack) {
long *first, *second;
popTwo(first, second, stack);
//Push the former first before the former second.
push(first, stack);
push(second, stack);
}
/* MODIFIES: standard output.
* EFFECTS: Prints the first element on the stack to standard output, followed
* by a newline.
*/
void print(Stack<long> &stack) {
long *operand;
pop(operand, stack);
printf("%li\n", *operand);
push(operand, stack);
}
/* MODIFIES: stack.
* EFFECTS: Empties the stack.
*/
void clear(Stack<long> &stack) {
while (!stack.isEmpty()) delete stack.pop();
}
/* MODIFIES: standard output.
* EFFECTS: Prints all the elements on the stack, from top to bottom, each
* separated by a single space. No trailing space. Ends with newline.
*/
void printAll(Stack<long> &stack) {
if (!stack.isEmpty()) {
const long *element = stack.begin();
do {
printf("%li", *element);
} while ((element = stack.next()) && printf(" "));
}
printf("\n");
}
/* MODIFIES: standard output.
* EFFECTS: Prints usage information and returns 0.
*/
int printUsage(const char* argv0) {
printf("Usage: %s [--help] [--usage] COMMAND ...\n", argv0);
printf("%s is a Reverse Polish Notation integer calculator.\n", argv0);
printf("\n");
printf(" --help, --usage Display this help and exit\n");
printf("\n");
printf("%s supports the following commands:\n", argv0);
printf("Any integers supplied are pushed onto the stack and can be given in decimal\n");
printf("(default), octal with a leading 0, or hexadecimal with a leading 0x. Numbers\n");
printf("can be prefixed with + or - to specify positive or negative, respectively.\n");
printf("The operators +, -, *, and / pop the top two elements and push the result of\n");
printf("<second> <operator> <first> to the stack. The top element is printed on exit\n");
printf("unless a print command is given or an error occurs. Also,\n");
printf(" a: prints all elements from first to last.\n");
printf(" c: clears the stack.\n");
printf(" d: duplicates the top element.\n");
printf(" n: negates the top element.\n");
printf(" p: prints the top element.\n");
printf(" r: reverses the order of the top two elements.\n");
printf("\n");
printf("Examples:\n");
printf(" %s 2 2 +\n", argv0);
printf(" %s 0xCAFE42 0777 *\n", argv0);
return 0;
}
int main(int argc, char *argv[]) {
//If no arguments, --usage, or --help.
if (argc == 1 ||
(argc > 1 && (!strcmp(argv[1], "--usage") ||
!strcmp(argv[1], "--help"))))
return printUsage(argv[0]);
Stack<long> stack;
bool printedAnything = false;
//Skip program name in arguments list.
for (int i = 1; i < argc; i++) {
char *endPtr;
//Allow input in decimal, octal, or hex.
long numberInput = strtol(argv[i], &endPtr, 0);
//Make sure the string is a number and not empty.
if (*argv[i] && !(*endPtr)) {
push(numberInput, stack);
continue;
}
/* Input was not a valid number, attempt to parse as operator.
* Anything other than one character is not a valid operator.
*/
if (strlen(argv[i]) == 1) {
switch (argv[i][0]) {
case '+':
//Add
apply(add, stack);
continue;
case '-':
//Subtract
apply(subtract, stack);
continue;
case '*':
//Multiply
apply(multiply, stack);
continue;
case '/':
//Divide
divide(stack);
continue;
case 'n':
//Negate
negate(stack);
continue;
case 'd':
//Duplicate
duplicate(stack);
continue;
case 'r':
//Reverse
reverse(stack);
continue;
case 'p':
//Print
print(stack);
printedAnything = true;
continue;
case 'c':
//Clear
clear(stack);
continue;
case 'a':
//Print-all
printAll(stack);
printedAnything = true;
continue;
}
}
printf("%s: unsupported command: %s\n", program_invocation_name, argv[i]);
exit(1);
}
if ( !printedAnything && !stack.isEmpty() ) {
print(stack);
}
return 0;
}