重温C语言(1)----计算算术表达式的值

《C程序设计语言》练习题 5-10

编写程序 expr，计算从命令行输入的逆波兰表达式的值，其中每个运算符或操作数用一个单独的参数表示。例如，命令

expr 2 3 4 + *

计算表达式2x（3+4）的值

算法实现

getfloat.c:

// getfloat.c
#include<stdio.h>
#include<ctype.h>
int getfloat(char* str, double* store)
{
	while (isspace(*str))
		str++;
	int ch, sign;
	double power = 1.0;
	ch = *str++;
	sign = ch == '-' ? -1 : 1;
	if (ch == '+' || ch == '-') {
		ch == *str++;
	}
	for (*store == 0.0; isdigit(ch); ch = *str++) {
		*store = *store*10.0 + ch - '0';
	}
	if (ch == '.') {
		ch=*str++;
	}
	for (; isdigit(ch); ch = *str++) {
		*store = *store*10.0 + ch - '0';
		power *= 10.0;
	}
	if (ch == 0 || isspace(ch) || ch == EOF) {
		//如果不能继续读取数字，那么跳出的原因必须是 ch==0 || isspace(ch) || ch==EOF
		*store /= power;
		*store *= sign;
		return 1;
	}
	else {
		return -1;
	}
}

expr.c:

//expr.c
#include<stdio.h>
#include<ctype.h>
#define MAX_NUM 100
double stack[MAX_NUM];
extern int getfloat(char* str,double*);
int main(int argc, char* argv[])
{
	if (argc < 2) {
		printf("usage: expr num1 num2 operator ...\n");
		return 1;
	}
	else {
		int stackp = 0;
		int ch = 0;
		double num1, num2, ans;
		while (--argc >= 1) {
			ch = (*++argv)[0];
			if (ch == '-'){
				if (!isdigit(ch = (*argv)[1])) {
					//是负号
					if (stackp >= 2) {
						num2 = stack[--stackp];
						num1 = stack[--stackp];
						stack[stackp++] = num1 - num2;
						continue;
					}
					else {
						printf("Error: at least two numbers are needed before operator '-'\n");
						return -1;
					}
				}
			}
			if (isdigit(ch)) {
				if (stackp < 0) {
					printf("stack is underflowed!\n");
					return -2;
				}
				else if (stackp >= MAX_NUM) {
					printf("stack is overflowed!\n");
					return -3;
				}
				else if (getfloat(*argv, &stack[stackp++]) == -1) {
					printf("wrong argument to get number: %s\n", *argv);
					return -4;
				}
			}
			else if (ch == '+'||ch=='*'||ch=='/'||ch=='%') {
				if (stackp >= 2) {
					num2 = stack[--stackp];
					num1 = stack[--stackp];
					switch (ch) {
					case '+':
						ans = num1 + num2;
						break;
					case '*':
						ans = num1*num2;
						break;
					case '/':
						if (num2 == 0.0) {
							printf("error zero divisor: %f / %f\n", num1, num2);
							return -5;
						}
						ans = num1 / num2;
						break;
					case '%':
						ans = (int)num1%(int)num2;
						break;
					}
					stack[stackp++] = ans;
				}
				else {
					printf("Error: at least two numbers are needed before operator '%c'\n",ch);
					return -6;
				}
			}
			else {
				printf("wrong arguments for %c,usage: expr num1 num2 operator ...\n", ch);
				return -7;
			}
		}
		if (stackp == 1) {
			printf("%.4f\n", stack[0]);
		}
		else {
			printf("the format of input is wrong!\n");
		}
	}
	return 0;
}