简单的计算器，在C中使用if开关

Question

This is my code so far for my simple calculator. Im working on sine right now(case 6) with a degree range of 0-360.Here is the output.

$ ./a.exe ProblemSolving.c
Arithmetic              : Add(0) Sub(1) Mult(2) Div(4) Mod(5)
Trigonometry    : sine(6) cosine(7) tan(8) arc_sin(9) arc_cos(10)
Exponent                : x^y(11) 2^x(12) 10^x(13)
Enter the choice of operation:6
The choice of operation is:6
Enter degree range from 0 to 360
Enter degrees:400

After I enter the desired degrees nothing else happens and the program ends. I believe there is something wrong with my if statement or the sine function.

#include <stdio.h>
#include <math.h>

int main() {        
    double Add1, Add2, Sub1, Sub2, Mult1, Mult2;
    int Choice, Div1, Div2, Mod1, Mod2, Base1, Power1, Deg1; 

    printf("Arithmetic      : Add(0) Sub(1) Mult(2) Div(4) Mod(5)\n");
    printf("Trigonometry    : sine(6) cosine(7) tan(8) arc_sin(9) arc_cos(10)\n");
    printf("Exponent        : x^y(11) 2^x(12) 10^x(13)\n");
    printf("Enter the choice of operation:");
    scanf("%d", &Choice);
    printf("The choice of operation is:%d\n", Choice);

    switch(Choice) {
            case 0:
                printf("Enter number one:");
                scanf("%lf", &Add1);
                printf("Enter number two:");
                scanf("%lf", &Add2);
                printf("%2.2lf + %2.2lf = %2.2lf", Add1, Add2, Add1+Add2);
                break;
            case 1:
                printf("Enter number one:");
                scanf("%lf", &Sub1);
                printf("Enter number two:");
                scanf("%lf", &Sub2);
                printf("%2.2lf - %2.2lf = %2.2lf", Sub1, Sub2, Sub1-Sub2);
                break;
            case 2:
                printf("Enter number one:");
                scanf("%lf", &Mult1);
                printf("Enter number two:");
                scanf("%lf", &Mult2);
                printf("%2.2lf * %2.2lf = %2.2lf", Mult1, Mult2, Mult1*Mult2);
                break;
            case 4:
                printf("Enter number one:");
                scanf("%d", &Div1);
                printf("Enter number two:");
                scanf("%d", &Div2);
                if (Div2 == 0)
                    printf("Error! Denominator cannot equal 0");
                else 
                    printf("%d / %d = %d", Div1, Div2, Div1/Div2);
                break;
            case 5:
                printf("Enter number one:");
                scanf("%d", Mod1);
                printf("Enter number two:");
                scanf("%d", Mod2);
                if (Mod2 == 0)
                    printf("Error! Denominator cannot equal 0");
                else
                    printf("%d % %d = %d", Mod1, Mod2, Mod1%Mod2);
                break;
            case 6:
                printf("Enter degree range from 0 to 360\n");
                printf("Enter degrees:");
                scanf("%d", Deg1);
                if (0 > Deg1 > 360) 
                    printf("Error! Value Entered is not within valid range");
                else 
                    printf("sin(%d) = %d", Deg1, sin(Deg1));
                break;
            default:
                printf("Error! operator is not correct");
                break;  
    }
    return 0;
}   

Answer 1

The sine function (and the rest of the trig functions) in C works in radians, not degrees. You'll have to convert from degrees from radians before passing the value to sine .

Right now you also have a problem with the format in your printf , since you're passing a double , but telling printf to expect an int . You need to use %f instead of %d .

In addition, your if statement doesn't currently make much sense, and almost certainly doesn't mean what you think. What you apparently want is if (Deg1 < 0.0 || Deg1 > 360.0)

Answer 2

There are several problems in this code:

1. Change scanf("%d", Deg1); to scanf("%d", &Deg1); , because scanf() call for addresses. Also, I think it might be better to declare Deg1 as a double .
2. 0 > Deg1 > 360 is wrong in C. You have to write Deg1 < 0 || Deg1 > 360 Deg1 < 0 || Deg1 > 360 . Operator || stands for "logical Or".
3. In math.h , sin() works in radians. So use sin(Deg1 * 3.14159265 / 180) . Or, to improve readability and maintenance, #define PI 3.14159265 and sin(Deg1 * PI / 180) . Note that you cannot write Deg1 / 180 * 3.14159265 , because integer literals in C is int s, and int / int = int . For example, 3 / 2 == 1, rather than 1.5. To get the exact value, write 3.0 / 2.0.
4. In math.h , sin() returns double , so write printf("sin(%d) = %g", Deg1, sin(...)); .

Fixed code here:

#include <stdio.h>
#include <math.h>

#define PI 3.14159265

// ...many lines of code...

    case 6:
        printf("Enter degree range from 0 to 360\n");
        printf("Enter degrees:");
        scanf("%d", &Deg1);
        if (0 > Deg1 || Deg1 > 360) 
            printf("Error! Value Entered is not within valid range");
        else 
            printf("sin(%d) = %g", Deg1, sin(Deg1 * PI / 180));
        break;