Thread: Help with simple function

Hi guys, first post so take it easy. I just started programming yesterday in C. And decided to try and write a simple programme to calculate the cos of a user input value. This is my attempt so far but its not working

Code:

#include <stdio.h>
#include <math.h>
main()
{
int d;
float x;
	printf("welcome to the cos calculator\n");
	printf("what angle do you want the cos of:\n");
	scanf("%d" &d);
	x=cos(d);
	printf("x=%0.2f\n",x);
}

I know this is probably completely wrong so any help would be much appreciated
Thanks

Code:

	scanf("%d", &d);

Btw, Read the documentation of all functions you used.google is your friend.

The cos() function returns the cosine of x, where x is given in radians.

The main function needs a type specifier, namely int:

Code:

int main()
{
    ...
    return 0;
}

Need a comma in that scanf statement:

Code:

scanf("%d", &d);

Pretty sure the cos function expects radians, so you'll need to convert from degrees to radians.

cos() expects a double. Conversion is implied but your compiler should have given you a warning when you passed it an int. It returns a double, so when you assign to a float there's another conversion implied. In the end it's probably easiest to only work with doubles.

Thanks guys, I got it to work with the following code, I then added a bit more on to see how much i could do. But i have 2 questions. Firstly why do you have to put "int" in front of "main ()" as we did not have to that for other stuff we learned in class.

Code:

#include <stdio.h>
#include <math.h>
int main()
{
int r;
double x;
double d;
	printf("welcome to the cos calculator\n");
	printf("what angle do you want the cos of):\n");
	scanf("%d", &r);
	d=r*0.01745;
	x=cos(d);
	printf("x=%0.5f\n",x);
int y;
printf("are you happy with the answer?(yes=1/no=0)\n");
scanf("%d",&y);
	if(y==1) {
	printf("huzzah!!\n");
	}
	else if(y==0){
	printf("Too bad, it's right!\n");
	}
}

and second of all, I had to use integers for the yes/no part at the end, because I couldn't figure out how to just use the input "y" or "n". How could I've done this instead?
Many thanks, alex

int main() vs. void main()

To read a single char, you can use getchar()

Thanks guys, I got it to work with the following code, I then added a bit more on to see how much i could do. But i have 2 questions. Firstly why do you have to put "int" in front of "main ()" as we did not have to that for other stuff we learned in class.

Code:

#include <stdio.h>
#include <math.h>
int main()
{
int r;
double x;
double d;
	printf("welcome to the cos calculator\n");
	printf("what angle do you want the cos of(in radians please):\n");
	scanf("%d", &r);
	d=r*0.01745;
	x=cos(d);
	printf("x=%0.5f\n",x);
int y;
printf("are you happy with the answer?(yes=1/no=0)\n");
scanf("%d",&y);
	if(y==1) {
	printf("huzzah!!\n");
	}
	else if(y==0){
	printf("Too bad, it's right!\n");
	}
}

and second of all, I had to use integers for the yes/no part at the end, because I couldn't figure out how to just use the input "y" or "n". How could I've done this instead?
Many thanks, alex

Originally Posted by alexl

Firstly why do you have to put "int" in front of "main ()" as we did not have to that for other stuff we learned in class.

Fire your teacher! Then, check out the link Bayint Naung provided, or check here. It's always best to be explicit and follow the standard, since you never know what compiler somebody else may be using.

and second of all, I had to use integers for the yes/no part at the end, because I couldn't figure out how to just use the input "y" or "n". How could I've done this instead?
Many thanks, alex

Code:

printf("are you happy with the answer?(yes=y/no=n)\n");
scanf("%c", &c);
if (c == 'y') {
    printf("huzzah!!\n");
}
else if(y == 'n') {
    printf("Too bad, it's right!\n");
}
else {
    printf("Can't you follow directions?  I said press 'y' or 'n'!\n");
}