c - 应为 ‘float *’，但参数的类型为 ‘float’，与 ‘find_m’ 的参数 2 的类型不兼容

Question

我正在尝试用 C 语言编写最小二乘法。

cc least_square.c -lm -o least_square

用 gcc 编译后出现此错误。

least_square.c: In function ‘main’:
least_square.c:37:32: error: incompatible type for argument 1 of ‘find_m’
           printf("m = %.3f\n", find_m(x[i], x[i], k)) ; 
                                ^
least_square.c:8:11: note: expected ‘float *’ but argument is of type ‘float’
     float find_m(float [], float [], int data_number) ;
           ^
least_square.c:37:32: error: incompatible type for argument 2 of ‘find_m’
           printf("m = %.3f\n", find_m(x[i], x[i], k)) ; 
                                ^
least_square.c:8:11: note: expected ‘float *’ but argument is of type ‘float’
     float find_m(float [], float [], int data_number) ;
           ^
least_square.c:38:32: error: incompatible type for argument 1 of ‘find_n’
           printf("n = %.3f\n", find_n(x[i], y[i], k)) ; 
                                ^
least_square.c:10:11: note: expected ‘float *’ but argument is of type ‘float’
     float find_n(float [], float [], int data_number) ;
           ^
least_square.c:38:32: error: incompatible type for argument 2 of ‘find_n’
           printf("n = %.3f\n", find_n(x[i], y[i], k)) ; 
                                ^
least_square.c:10:11: note: expected ‘float *’ but argument is of type ‘float’
     float find_n(float [], float [], int data_number) ;

我的代码: /* y = mx + n --- 使用最小二乘法 */

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


    float find_m(float [], float [], int data_number) ;

    float find_n(float [], float [], int data_number) ;


    int main(){


       int k, i ;    

       printf("Number of data(k)\n? ") ;
       scanf("%d", &k) ;

       float x[k] ;    
       float y[k] ;

       for (i = 0 ; i <= (k - 1) ; i++){    
          x[i] = 0 ;
          y[i] = 0 ; 
       } 

       for (i = 0 ; i <= (k - 1) ; i++){

          printf("%d. point x coordinate(xi)\n? ", i+1) ;  /* get data */
          scanf("%f", &x[i]) ;

          printf("%d. point y coordinate(yi)\n? ", i+1) ;
          scanf("%f", &y[i]) ;

          printf("m = %.3f\n", find_m(x[i], x[i], k)) ; 
          printf("n = %.3f\n", find_n(x[i], y[i], k)) ; 
       }


       return 0 ;       


    }


    float find_m(float x[], float y[], int data_number){   /* Find constant m */


       int i,
           a = 0, 
           b = 0, 
           c = 0,
           d = 0,
           e = 0 ; 

       for (i = 0 ; i <= (data_number - 1) ; i++){

          a += x[i] * y[i] ;

          b += x[i] ;

          c += y[i] ;

          d += pow(x[i], 2) ;

          e += x[i] ; 
       }   


       return ((data_number * a) - (b * c)) / ((data_number * d) - pow(e, 2)) ;


    }


    float find_n(float x[], float y[], int data_number){        /*Find constant n*/


       int i,
           a = 0, 
           b = 0, 
           c = 0,
           d = 0,
           e = 0,
           f = 0;
       float n ; 

       for (i = 0 ; i <= (data_number - 1) ; i++){

          a += pow(x[i], 2) ;

          b += y[i] ;

          c += x[i] * y[i] ;

          d += x[i] ;

          e += pow(x[i], 2) ;

          f += x[i] ; 
       }   


       return (((a * b) - (c * d)) / ((data_number * e) - pow(f, 2))) ;


    }

我无法弄清楚我的错误是什么。我认为一切都是对的。

错误提示“expected float *”，但我已经将参数类型设置为 float。

Answer 1

find_m 和 find_n 都有指向 float 的指针作为前两个参数。他们的签名相当于:

float find_m(float*, float*, int data_number) ;

float find_n(float*, float*, int data_number) ;

您正在将 float 作为参数传递。您需要传递指针。如果不完全理解代码的用途，就很难提供真正有效的解决方案。一种可能性是您希望在每次迭代中扫描部分数组，在这种情况下，您可以使用数组名称作为参数。这些将衰减为指向数组第一个元素的指针。

printf("m = %.3f\n", find_m(x, x, k)) ; 
printf("n = %.3f\n", find_n(x, y, k)) ;