Question

我正在使用C进行练习，我将数据输入到结构中，然后在单独的函数中对其进行操作。但是当程序进入实际数学运算的行时，我得到一个关于被调用对象不是函数的错误。

这是确切的错误：

p1s2.c:70:106: error: called object '(vectorArray + (sizetype)((unsigned int)i * 32u))->y * (vectorArray + (sizetype)((unsigned int)i * 32u))->y' is not a function

我会提前为代码道歉，这是一个正在进行中的文件，所以它不是很干净。

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
//Structure declaration.
struct vector {
    double x;               //X-coordinate for vector
    double y;               //Y-coordinate for vector
    double z;               //Z-Coordinate for vector
    double length;      //Length. Calculations will go here.
};

int howLong(struct vector *x);
int main(void)
{
    int arraySize;
    int i;
    int vectorNum=1;
    int retval;                 //Watch for counter in howLong.
    int scanval;            //Error checking for coordinates input (scanf statement)

    printf("How many vectors would you like to calculate length for?\n");
    scanf("%d", &arraySize);
    //Allocate memory to struct.
    struct vector *vectorArray = malloc(arraySize*sizeof(double));

    printf("You will now enter the coordinates for %d vectors. \n", arraySize); 
    //Input loop.
    for(i=0; i<=arraySize; i++){
        printf("Please enter the X, Y, Z coordinates for vector %d. \n", vectorNum);
        printf("Please separate the coordinates with spaces. \n");

        int scanval;            //Error checking: Scanval should be equal to three. 
        //scanf takes user input, converts to long float.
        scanval=(scanf("%lf %lf %lf", &vectorArray[i].x, &vectorArray[i].y, &vectorArray[i].z));
        if(scanval !=3) {
            printf("You can't follow directions. That's too bad. \n");
            exit(0);
        }
        //Print input back to user.
        printf("Vector Number %d: %lf %lf %lf \n", vectorNum, vectorArray[i].x, vectorArray[i].y, vectorArray[i].z);
        //Increment counters.
        vectorNum++;
        i++;
    }
    for(i=0; i<=arraySize; i++){
        vectorArray[i].length=howLong(vectorArray);
        i++;

    }   
}

这是外部函数howLong：

 int howLong(struct vector *vectorArray) 
    { //Function gets the struct and coordinate values, calculates length and writes to struct.
        int i;
        int vectorNum=1;
        printf("Calculating vector length. \n"); 
        //Math.
        vectorArray[i].length = sqrt((vectorArray[i].x * vectorArray[i].x)+(vectorArray[i].y * vectorArray[i].y)(vectorArray[i].z * vectorArray[i].z));
 //Error occurs on the line directly above this comment. ^^     

printf("Length of Vector Number %d: %lf \n", vectorNum, vectorArray[i].length);
        return vectorArray[i].length; 
    }

我不明白。起初我认为它与函数名称有关，但在我将函数名称更改为howLong后错误仍然存在。有任何想法吗？

Answer 1

嗯，这里：

(vectorArray[i].y * vectorArray[i].y)(vectorArray[i].z * vectorArray[i].z)

在两个（）部分之间缺少加号。

Answer 2

在函数howLong

中

((vectorArray[i].x * vectorArray[i].x)+(vectorArray[i].y * vectorArray[i].y)(vectorArray[i].z * vectorArray[i].z));        
                                                                            ^operator is missing

公式抛出错误'被叫对象不是一个函数'

2 个答案: