錯誤：預計'，'或';'在'{'標記之前

Question

我確定有一個非常簡單的答案，但我已經查找了幾個小時，但無法弄清楚。正如標題所說，我收到了一個我在代碼中評論過的錯誤。一切似乎都是用正確的語法，所以我不知道它會是什麼。

#include <iostream> 
#include <math.h> 
using namespace std; 
float x, y, k, q, base, exponent; 

/*float atan(q) 
{ 
    //I had made my own functions for all the different math functions that 
    would have normally been included in the math.h file, but I had no idea 
    how to tackle making a function for an arc tangent as all the information 
    I could find online just had you kinda guess what the end result might be, 
    so I used the math.h file for this and only this function. 
}*/ 
float sqrot(k) 
{ // <----- the error is here 
    float guess, divide, average, z; 
    guess=rand()%k; 
    for(z;z<500;z++) 
    { 
     divide=k/guess; 
     average=(guess+divide)/2; 
     guess=average; 
    } 
} 
float cosine(y) 
{ 
    y=1-(((y*y)/2)+((y*y*y*y)/24)-((y*y*y*y*y*y)/720)); 
    return y; 
} 
float sine(x) 
{ 
    x=x-(((x*x*x)/6)+((x*x*x*x*x)/120)-((x*x*x*x*x*x*x)/5040)); 
    return x; 
} 
float power(base, exponent) 
{ 
    while(exponent>1) 
    { 
     base=base*base; 
     exponent-1; 
    } 
    return base; 
} 
float haversine(lat1,long1,lat2,long2) 
{ 
    float degree_to_rad, pi=3.14159; 
    int d_lat, d_long, a, c, mi; 
    degree_to_rad=pi/180; 
    d_lat=(lat2-lat1)*degree_to_rad; 
    d_long=(long2-long1)*degree_to_rad; 
    a=power(sine(d_lat/2),2)+cosine(lat*degree_to_rad)*cosine(lat2*degree_to_rad)*power(sine(d_long/2),2); 
    c=2*atan((sqrot(a))/(sqrot(1-a))); 
    mi=3956*c; 
    return mi; 
} 
int main() 
{ 
    int answer; 
    cout<<"Enter the Latitude of your starting location: "; 
    cin>>lat1; 
    cout<<"Enter the Longitude of your starting location: "; 
    cin>>long1; 
    cout<<"Enter the Latitude of your ending location: "; 
    cin>>lat2; 
    cout<<"Enter the Longitude of your ending location: "; 
    cin>>long2; 
    answer=haversine(lat1, long1, lat2, long2); 
    cout<<"The distance between the two points is: "<< answer; 
    return 0; 
} 

Answer 1

正式參數必須具有類型

float sqrot(k) 

應該已經

float sqrot(type k) 
     // ^^^^ 

，同樣與