Program to emulate calculator

**crucified** · 09-02-2008

Hi all,

i was told to do up a program to emulate a calculator (+-*/ functions) but i am unable to get the program to show my answer.
attached is the program i typed. can someone help me out?

Code:

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

double Addition(double num1, double num2){
    return num1+num2;
}
double Subtraction(double num1, double num2){
    return num1-num2;
}
double Multiplication(double num1, double num2){
    return num1*num2;
}
double Division(double num1, double num2){
    return num1/num2;
}
main (int argc, char *argv[]){
    char sign;
    double first_num, second_num;
    printf("Type 2 numbers and an operation\n");
    scanf("%lf %c %lf", &first_num, &sign, &second_num);

    if (&sign == "+")
        printf("%f\n", Addition(first_num, second_num));
 else if (&sign == "-")
        printf("%f\n", Subtraction(first_num, second_num));
 else if (&sign == "*")
        printf("%f\n", Multiplication(first_num, second_num));
 else if (&sign == "/")
        printf("%f\n", Division(first_num, second_num));
}

**QuantumPete** · 09-03-2008

When you do (&sign == "+") you're actually performing an address comparison. You're comparing the address of the variable sign (which you get by prepending the &) with the address of the global string literal "+".
What you really want is a character comparison. So you need to take the variable itself and compare agains the *character* +, which you do by using single quotes.
Also two other things to note are that you are missing the curly braces on your if and else if statements and that the format specifier in your printf statements is missing the l (ell) that makes it print out a double instead of a float.

QuantumPete

**matsp** · 09-03-2008

Originally Posted by QuantumPete

that the format specifier in your printf statements is missing the l (ell) that makes it print out a double instead of a float.

Whilst the rest of your post is indeed correct, this part is not.
1. All floating point values are promoted to double when passed to var-args functions (and functions with unspecified arguements - old-style code). Thus %f is used for float and double - since they are all double anyways.
2. %lf is used in some compilers (c99) to indicate "long double", which would lead to erroneous output. In other compilers, %lf _may_ mean double, but in others it will be "unknown" [and thus in the clearly undefined behaviour arena].

--
Mats

**crucified** · 09-03-2008

it finally works!
haha thanks to the both of u pete and mats!

**QuantumPete** · 09-03-2008

Originally Posted by matsp

Whilst the rest of your post is indeed correct, this part is not.
1. All floating point values are promoted to double when passed to var-args functions (and functions with unspecified arguements - old-style code). Thus %f is used for float and double - since they are all double anyways.
2. %lf is used in some compilers (c99) to indicate "long double", which would lead to erroneous output. In other compilers, %lf _may_ mean double, but in others it will be "unknown" [and thus in the clearly undefined behaviour arena].

--
Mats

You live and learn. I was sure thaf %f is float and %lf is double...

**tabstop** · 09-03-2008

Originally Posted by QuantumPete

You live and learn. I was sure thaf %f is float and %lf is double...

For scanf, it is. For printf, it isn't.

**Elysia** · 09-03-2008

Originally Posted by QuantumPete

Also two other things to note are that you are missing the curly braces on your if and else if statements...

Oh, but it's not required. It's optional.
More like it's recommended. But then again, perhaps not, because it takes space.
There's no easy answer to that question, but it certainly isn't required.

Oh and you're using implicit main. That's bad.

**QuantumPete** · 09-03-2008

Originally Posted by Elysia

Oh, but it's not required. It's optional.

True, but it's better to have them and not need them, then to need them and not have them

QuantumPete

**QuantumPete** · 09-03-2008

Originally Posted by tabstop

For scanf, it is. For printf, it isn't.

[sarcasm]Well of course, that makes sense, two functions from the same library using the same format specifier meaning different things. [/sarcasm] There are days I hate C

QuantumPete

**vart** · 09-03-2008

same format specifier meaning different things

Not much sense to do different format specifiers for float and double in printf while both are passed as double

scanf is other story...

**Zihui02** · 09-04-2008

Hi, I also need help in the code... after trying out with the modified code, i still receive error upon running... and also unable to get program to show the answer
Can anyone help? Need it to be done urgently... ):

Code:

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

double Addition(double num1, double num2){
    return num1+num2;
}
double Subtraction(double num1, double num2){
    return num1-num2;
}
double Multiplication(double num1, double num2){
    return num1*num2;
}
double Division(double num1, double num2){
    return num1/num2;
}
main (int argc, char *argv[]){
    char operator;
    double first_num, second_num;
    printf("Type 2 numbers and an operator\n");
    scanf("&#37;f %c %f", &first_num, &operator, &second_num);

    if (&operator == '+') {
        printf("%f\n", Addition(first_num, second_num));
 }else if (&operator == '-') {
        printf("%f\n", Subtraction(first_num, second_num));
 }else if (&operator == '*'){
        printf("%f\n", Multiplication(first_num, second_num));
 }else if (&operator == '/'){
        printf("%f\n", Division(first_num, second_num));
}
}

**tabstop** · 09-04-2008

And you didn't read Pete's first response because why?

**Zihui02** · 09-04-2008

eh it's because I don't really understand.. i'm totally new to this..
and i did add in the brackets to if and if else statements, changing the " " to ' ' , %If to %f
it still don't work..

**tabstop** · 09-04-2008

And if you do the other bit mentioned, namely getting rid of &, what happens?

And if you don't understand what he said, look up the words in the index of your textbook to get a brush-up on the vocabulary.

**Elysia** · 09-05-2008

The & operator takes the ADDRESS of something and the ADDRESS of something generates an extra indirection, ie char becomes char*, and char* is not the same as char. The end.
And you still haven't fixed main either.
But do say, do you have a book or is it a course from where you're teaching yourself?

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