Too many arguments to function call, What do I do?

Question

I am working on a problem in my textbook and I need to make a triangle angle calculator Im used to java but Im not 100% on C yet, I don't understand the logic in it.

#include <stdio.h>

static float angleB;
static float angleA;

float remainingAngle(float answer)
{
    float answer= angleA+angleB;
//redefinition of answer
 return 0;
}
//CANT TOUCH FROM HERE TO END
int main(int argc, const char * argv[]) {
    float angleA = 30.0;
    float angleB = 60.0;
    float angleC = remainingAngle(angleA,angleB);
// to many arguments to function call (referring to angleC)
    printf("the third angle is %.2f\n",angleC);
    return 0;
}     //END CANT TOUCH

I don't know what to do here.

Answer 1

In your function definition

float remainingAngle(float answer)

the function remainingAngle() accepts one parameter.

OTOH, in your function call

remainingAngle(angleA,angleB);

you're passing two arguments.

The supplied argument number and type should match with the parameter list in function definition.

---

That said, your code is wrong.

• Point 1. Your local variables will shadow the global ones. Maybe that's not what you want. Change

  float angleA = 30.0;
  float angleB = 60.0;
  

  to

   angleA = 30.0;
   angleB = 60.0;
  

• Point 2. The function

  float remainingAngle(float answer)
  {
      float answer= angleA+angleB;
  //redefinition of answer
   return 0;
  }
  

  is logically wrong. It should rather be

  float remainingAngle(void)
  {
      float answer= angleA+angleB;
      return answer;
  }
  

  and should be called like

  float angleC = remainingAngle();
  

---

AFTER EDIT:

as per your requirement, you can do

float remainingAngle(float angleA, float angleB)
{
    float answer= angleA+angleB;
    return answer;
}

However, this makes the global variables useless.