Solve Exercise 5-10: expr

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diff --git a/ch5/5-10_expr.c b/ch5/5-10_expr.c
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+#include <stdio.h>
+#include <ctype.h>
+#include <stdlib.h>
+
+/* The C Programming Language: 2nd Edition
+ *
+ * Exercise 5-10: Write the program expr, which evaluates a reverse Polish
+ * expression from the command line, where each operator or operand is a
+ * separate argument. For example:
+ *
+ * expr 2 3 4 + *
+ *
+ * evaluates 2 × (3+4).
+ */
+
+#define STACKSIZE 100
+#define NUMBER '0'
+
+double stack[STACKSIZE];
+double *stp = stack; // stack pointer
+
+void push(double);
+double pop(void);
+
+/* This solution only offers basic arithmetic as proof of concept. See
+ * Chapter 4's exercises on how to extend this to a fully featured
+ * calculator. Note that for most shells, you'll need to escape the
+ * multiplication symbol (*) or it will expand to every file in the current
+ * directory; NOT what you want in this situation. :)
+ * 
+ * So if you want 6 × 8 for example, you'll need to type `6 8 \*`
+ */
+int main(int argc, char *argv[]) {
+	int argp; // current argument position
+	double op2;
+	for (argp = 1; argp < argc; argp++) {
+		printf("checking arg: %s\n", argv[argp]);
+		int argtype = opt_type(argv[argp]);
+		switch(argtype) {
+			case '*':
+				push(pop() * pop());
+				break;
+			case '+':
+				push(pop() + pop());
+				break;
+			case '-':
+				op2 = pop();
+				push(pop() - op2);
+				break;
+			case '/':
+				op2 = pop();
+				if (op2 != 0.0) {
+					push(pop() / op2);
+				} else {
+					printf("error: divide by zero\n");
+				}
+				break;
+			case NUMBER:
+				printf("it's a number\n");
+				push(atof(argv[argp]));
+				break;
+		}
+		printf("%s\n", argv[argp]);
+	}
+	printf("%f\n", pop());
+}
+
+void push(double num) {
+	if (stp >= stack + STACKSIZE) {
+		printf("Cannot push: stack full.\n");
+		return;
+	}
+	printf("add %f to stack.\n", num);
+	*(++stp) = num;
+}
+
+double pop(void) {
+	if (stp == stack) {
+		printf("Cannot pop: stack empty.\n");
+		return 0.0;
+	}
+	return *(stp--);
+}
+
+int opt_type(char op[]) {
+	int i = 0;
+	switch(op[i]) {
+		case '*':
+			return '*';
+		case '+':
+			return '+';
+		case '/':
+			return '/';
+		case '-':
+			if (op[i+1] == '\0') {
+				return '-';
+			}
+			break;
+	}
+
+	while (isdigit(op[i])) {
+		i++;
+	}
+	if (op[i] == '\0') {
+		return NUMBER;
+	}
+}