'f'
创建用户可以输入或'i'
在整数和浮点之间切换的浮点模式的最有效方法是什么?我想这样做而不必复制花车的整个代码。我知道类型转换是一种选择,但我不完全确定这是否是最安全的方式。
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define MAX 100
int *p;
int *tos;
int *bos;
void push(int i);
int pop(void);
int main (void)
{
int a, b;
char s[80];
p = (int *) malloc(MAX*sizeof(int)); /* get stack memory */
if (!p) {
printf("Allocation Failure\n");
exit(1);
}
tos = p;
bos = p + MAX-1;
printf("\nRPN Calculator\n");
printf("Enter 'i' for integer mode\n");
printf("Enter 'f' for floating point mode\n");
printf("Enter 'q' to quit\n\n");
char *endptr;
do {
printf("> ");
scanf("%s", s);
int val = strtol(s, &endptr, 10);
if (*endptr == '\0') {
//printf("Got only the integer: %d\n", val);
}
else {
printf("operator: %s\n", endptr);
printf("integer: %d\n", val);
if (val != 0){ /* don't push val on stack if 0 */
push(val);
}
}
switch(*endptr) {
case 'i':
printf("(Integer Mode)\n");
break;
case 'f':
printf("(Floating Point Mode)\n");
break;
case '+':
a = pop();
b = pop();
// printf("%d\n",a);
// printf("%d\n",b);
// printf("%d\n",val);
printf("%d\n", a+b);
push(a+b);
break;
case '-':
a = pop();
b = pop();
printf("%d\n", b-a);
push(b-a);
break;
case '*':
a = pop();
b = pop();
printf("%d\n", a*b);
push(a*b);
break;
case '/':
a = pop();
b = pop();
if(a == 0){
printf("Cannot divide by zero\n");
break;
}
printf("%d\n", b/a);
push(b/a);
break;
case '.':
a = pop(); push(a);
printf("Current value on top of stack: %d\n", a);
break;
default:
// push(atoi(s));
push(val);
}
} while (*s != 'q'); /* Do until 'q' is entered */
return 0;
}
void push (int i) /* Put an element on the stack */
{
if (p > bos){
printf("Stack Full\n");
return;
}
*p = i;
p++;
}
int pop (void) /* Get the element from the top of the stack */
{
p--;
if(p < 0) {
printf("Stack Underflow\n");
return 0;
}
return *p;
}