我正在用 C 做家庭作业。我必须构建一个计算器来接收 RPN，将其转换为双精度数，从堆栈中添加/删除它并打印堆栈上剩余的内容。在我运行它之前，一切似乎都很顺利。输入我的输入后，char[] 成功转换为 double（或者我认为是这样），并且在此过程中的某个地方（可能在 getop() 中）程序是否认为我没有输入数字。这是代码和输出。
#include <stdio.h>
#define MAXOP 100 /* maximum size of the operand of operator */
#define NUMBER '0' /* signal that a number was found */
int getOp(char []); /* takes a character string as an input */
void push(double);
double pop(void);
double asciiToFloat(char []);
/* reverse polish calculator */
int main()
{
int type;
double op2;
char s[MAXOP];
printf("Please enter Polish Notation or ^d to quit.\n");
while ((type = getOp(s)) != EOF) {
switch (type) {
case NUMBER:
push(asciiToFloat(s));
break;
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 : zero divisor!\n");
break;
case '\n':
printf("\t%.2f\n", pop()); /* this will print the last item on the stack */
printf("Please enter Polish Notation or ^d to quit.\n"); /* re-prompt the user for further calculation */
break;
default:
printf("error: unknown command %s.\n", s);
break;
}
}
return 0;
}
#include <ctype.h>
/* asciiToFloat: this will take ASCII input and convert it to a double */
double asciiToFloat(char s[])
{
double val;
double power;
int i;
int sign;
for (i = 0; isspace(s[i]); i++) /* gets rid of any whitespace */
;
sign = (s[i] == '-') ? -1 : 1; /* sets the sign of the input */
if (s[i] == '+' || s[i] == '-') /* excludes operands, plus and minus */
i++;
for (val = 0.0; isdigit(s[i]); i++)
val = 10.0 * val + (s[i] - '0');
if (s[i] = '.')
i++;
for (power = 1.0; isdigit(s[i]); i++) {
val = 10.0 * val + (s[i] - '0');
power *= 10.0;
}
return sign * val / power;
}
#define MAXVAL 100 /* maximum depth of value stack */
int sp = 0; /* next free stack position */
double val[MAXVAL]; /* value stack */
/* push: push f onto value stack */
void push(double f)
{
if (sp < MAXVAL) {
val[sp++] = f; /* take the input from the user and add it to the stack */
printf("The value of the stack position is %d\n", sp);
}
else
printf("error: stack full, cant push %g\n", f);
}
/* pop: pop and return the top value from the stack */
double pop(void)
{
if (sp > 0)
return val[--sp];
else {
printf("error: stack empty\n");
return 0.0;
}
}
#include <ctype.h>
int getch(void);
void ungetch(int);
/* getOp: get next operator or numeric operand */
int getOp(char s[])
{
int i;
int c;
while ((s[0] = c = getch()) == ' ' || c == '\t')
;
s[1] = '\0';
if (!isdigit(c) && c != '.') {
printf("Neither a digit or a decimal.\n");
return c; /* neither a digit nor a decimal */
}
i = 0;
if (isdigit(c)) /* grab the integer */
while (isdigit(s[++i] = c = getch()))
;
if (c == '.') /* grab the fraction */
while (isdigit(s[++i] = c = getch()))
;
s[i] = '\0';
if (c != EOF)
ungetch(c);
return NUMBER;
}
#define BUFSIZE 100
char buf[BUFSIZE]; /* buffer for ungetch */
int bufp = 0; /* next free position in buffer */
/* getch: get a number that may or may not have been pushed back */
int getch(void)
{
return (bufp > 0) ? buf[--bufp] : getchar();
}
/* ungetch: if we read past the number, we can push it back onto input buffer */
void ungetch(int c)
{
if (bufp >= BUFSIZE)
printf("ungetch: to many characters.\n");
else
buf[bufp++] = c;
}
输出 ：
请输入波兰表示法或 ^d 退出。123 堆栈位置的值为 1
既不是数字也不是小数。
123.00 请输入波兰表示法或 ^d 退出。
任何关于正在发生的事情的想法都会非常有帮助。似乎数字被正确传递，从char正确格式化为double，然后出现问题。
谢谢你。
岩石