Thread: An Exercise i am unable to complete fully, Please help.

Hey there, im currently trying to learn c and im unable to complete this own program,

"write a program that acts as a simple "printing" calculator. this program should allow the user to type in expressions of the form
number operator
the following operators should be recognised by program:
+, -, *, /, S, E

the S operator tells the program to set the accumulator to the typed in number. the E operator tells the program that execution is to end. The arithmetic operations are performed on the contents of the accumulator with the number that was keyed in acting as the second operand. The following is a "sample run" showing how the program should operate:

begin calculations
10 S //set accumulator to 10
=10.000 //contents of accumulator
2 / //divide by 2
=5.000 //contents of accumulator
55 - //subtract 55
=-50.000
100.25 S //set accumulator to 100.25
=100.250
4 * //multiply by 4
=401.000
0 E //end of program
=401.00
end of calculations

make certain that the program detects division by zero and also checks for unknown operators."

thank you .

this isnt a homework just an exercise from a programming in c book by stephen G. kochan. trying to learn how to code for my university

you input those lines in continuously order after each output thats where im stuck, i get the program to work but im allowed to input once and the program ends automatically after an the 1st output

A fuzzy picture is a poor substitute to posting the code itself as text. Remember to put it in code tags.

Code:

 
#include <stdio.h>
#include <stdlib.h>


int main ()
{

    float value1, A;
    char operator;
    

    printf("Enter in an expression: \n");
    scanf("%f %c ", &value1, &operator);
    
    
    
    switch(operator)
{
    
    case 'S':
                A = value1;
                printf("%.3f\n", A);
                break;
    
    case '+':
                printf("%.3f\n", value1 + A);
                break;
    case '-':
                printf("%.3f\n", value1 - A);
                break;
    case '*':
                printf("%.3f\n", value1 * A);
                break;
    case '/': 
                if (value1 == 0)
                printf("Error not possible, please try again\n");
                
                else 
                printf("%.3f\n", value1 / A);
                break;
                
    case 'E':     
                exit(1);
        
    
    
    default:
                printf("Unknown operator\n");
                break;
}                

        
                return 0;
}

You need a loop around pretty much the whole thing.

Code:

while (1) {
    printf();
    scanf();
    switch() {
    }
}

You should get rid of the extra space at the end of the scanf format. It will cause the input to hang there.

You need to "accumulate" the answer, so you need to set A to the new value every time. You could probably print the current value in one place, after the switch.

You may as well allow both upper- and lower-case S and E.

Code:

case 'E': case 'e':
    exit(0);

An exit value (or return value from main) of non-zero generally indicates program failure.

thank you it worked, appreciate everyones help