评估C中的表达式（例如1 + 2.5 * 3）

Question

This is a problem that i encountered while learning to program in c by following this book.

Enter an expression : 1+2.5*3
Output: 10.5

This is what i came up with so far. EDIT

#include<stdio.h>

int main (void)

{
char c;

float f1=0.0f,f2=0.0f;

c = getchar();
while(c != '\n')

{

  if(c!='+' && c!='-' && c!='*' && c!='/')
      {
          if (f1 == 0.0f)
          {
              f1 = c - '0';
              c = getchar();
          }
          else
          {
              f2 = c - '0';
              c = getchar();
          }

      }

 switch (c)
      {
          case '+': c = getchar(); f2 = c - '0' ; f1 = f1+f2; break;
          case '-': c = getchar(); f2 = c - '0' ; f1 = f1-f2; break;
          case '*': c = getchar(); f2 = c - '0' ; f1 = f1*f2; break;
          case '/': c = getchar(); f2 = c - '0' ; f1 = f1/f2; break;
         // case '.':
          default: break;
      }

 c = getchar();

 }
 printf("\n Value of the expression: %.2f", f1);
 return 0;
 }

The code is only implimented in assuming that all of the operands will be one digit number. How do i impliment it for more than one digit/floating point number? What approach should i take to solve this problem.

I have no instructor to consult with(self learning & at chapter 7) and i am stuck at this for hours.So any help will be hugely appreciated.

Thank You

NB

someone mentioned about atoi() ... but I am looking for something else/manual

Answer 1

f2 = c; // this will convert ascii charecter value of c into f2. 
         // How do i convert the character '2' into the digit 2 ?

This way:

f2 = c - '0';

Characters '0' to '9' are guaranteed to have sequential values in C.

Answer 2

If you're working in 'C', to convert a text representation of a "number" to the actual number -regardless of length- use something like atoi(). The "shortcut" that's possible when converting only one character isn't expandable or generalizable, and will have to be removed eventually. With something like atoi() on the other hand you can move from single-character numbers to multiple-character numbers without having to change any code.

The atoi() system library code has been around for decades and completely handles all the edge cases (negative zero anyone?); trying to roll your own would take an awful lot of effort to produce equivalently crisp results.