到目前为止,这是我的简单计算器代码。我正在研究sine
(案例6),学位范围为0-360。这是输出。
$ ./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
在我输入所需的度数后,没有其他事情发生,程序结束。我认为我的if
声明或sine
函数存在问题。
#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;
}
答案 0 :(得分:1)
C中的sine
函数(以及其余三角函数)以弧度而非度数工作。在将值传递给sine
之前,您必须从弧度转换度数。
目前您的printf
格式也存在问题,因为您传递了double
,但告诉printf
期待int
。您需要使用%f
代替%d
。
此外,您的if
声明目前没有多大意义,几乎可以肯定并不代表您的想法。您显然想要的是if (Deg1 < 0.0 || Deg1 > 360.0)
答案 1 :(得分:1)
此代码中存在以下几个问题:
scanf("%d", Deg1);
更改为scanf("%d", &Deg1);
,因为scanf()
要求地址。另外,我认为将Deg1
声明为double
。0 > Deg1 > 360
错了。你必须写Deg1 < 0 || Deg1 > 360
。运营商||
代表“逻辑或”。math.h
中,sin()
以弧度为单位。所以使用sin(Deg1 * 3.14159265 / 180)
。或者,为了提高可读性和维护性,#define PI 3.14159265
和sin(Deg1 * PI / 180)
。请注意,您不能编写Deg1 / 180 * 3.14159265
,因为C中的整数文字是int
s,而int
/ int
= int
。例如,3/2 == 1,而不是1.5。要获得准确的值,请写入3.0 / 2.0。math.h
,sin()
返回double
,所以请写printf("sin(%d) = %g", Deg1, sin(...));
。此处修正了代码:
#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;