錯誤：「computeTemp」的衝突類型（C）

Question

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

double computeTemp(double, double, double, double, double, double); 

void main(void) 
{ 

double rm,deg,h,w,l,t; 
    double final_temp; 

    printf("For all calculations assume that pressure is constant at 101.325kPa,and the heat capacity of air is also constant at 0.718kJ/kg*K\n"); 
    printf("Please enter the number of students in the room:\n"); 
    scanf("%lf",&rm); 
    printf("Please enter a temperature in degrees Celcius:\n"); 
    scanf("%lf",&deg); 
    printf("Please enter the height of the room in metres (m):\n"); 
    scanf("%lf",&h); 
    printf("Please enter the width of the room in metres (m):\n"); 
    scanf("%lf",&w); 
    printf("Please enter the length of the room in metres (m):\n"); 
    scanf("%lf",&l); 
    printf("Please enter a time in minutes (min.):\n"); 
    scanf("%lf",&t); 

    final_temp = computeTemp(rm,deg,h,w,l,t,final_temp); 

    printf("Number of students: %f\n",rm); 
    printf("Initial temperature in degrees Celcius: %f\n",deg); 
    printf("Height of room (m): %f\n",h); 
    printf("Width of room (m): %f\n",w); 
    printf("Length of room (m): %f\n",l); 
    printf("Time elapsed (min.):%f\n",t); 
    printf("Final Temperature in degrees Celcius: %f\n",final_temp); 

} 

double computeTemp(double num1, double num2, double num3, double num4, double num5, double num6) 

{ 

double temp; 

    temp = ((101.325*28.97)*(num3*num4*num5))/8.314*(num2+273.15); 
    temp = temp*0.718; 
    temp = (((4.8*num1*num6)*(num2+273.15))/(temp)); 
    temp = temp+(num2+273.15); 
    temp = temp-273.15; 
    return(temp); 
} 

我從函數調用刪除了一個double，並且出現了computeTemp參數太少的錯誤。我是新來的C語言編程，並使用代碼塊，這樣可以幫助我解決這個問題的任何信息將是非常讚賞

Answer 1

你向前聲明聲明computeTemp作爲

double computeTemp(double, double, double, double, double, double, double); 

和定義

double computeTemp(double num1, double num2, double num3, double num4, double num5, double num6) 

數它們;第一個有七個參數，而第二個只有六個...