Question

我的sqrt（）函数无效，我从命令行编译并追加-lm

/** initialize distance formula **/
double formula1 = (x2-x1) * 2 * 2; 
double formula2 = (y2-y1) * 2 * 2;
double formula = formula1 + formula2;
/** call sqrt function to square root it **/
sqrt(formula);
printf("\n%lf", sqrt(formula));

return sqrt(formula);

我的输出答案是-nanSTUDENTID @ gio：〜$

在这里编辑我的完整代码，我道歉。

#include <stdio.h>
/** call math.h libbrary cause of sqrt() **/
#include <math.h>

/** define/declare constant **/
#define PI 3.14159

/** declaring prototypes **/
double distance(double x1, double x2, double y1, double y2);
double radius(double point1, double point2, double point3, double point4);
double circumference(double circle_circum);
double area(double circle_area);

/** declare pointer **/
void getXY(double *xPtr, double *yPtr);

int main (void) {
    /** neither variables are initialized **/
    double x;
    double y;
    /** variables are now initialized **/
    getXY(&x, &y);

}

/** this function will prompt the user to enter x and y coordinates twice and then pass by reference to distance **/
void getXY(double *xPtr, double *yPtr) {
    /** neither variables are initialized **/
    double a;
    double b;
    /** variables are now initialized **/
    *xPtr = a;
    *yPtr = b;

    printf("First, lets get the center of a circle. \nPlease enter the x and y coordinates, separated by a space: \n" );
    scanf("%lf %lf", &a, &b);

    /** neither variables are initialized **/
    double c;
    double d;
    /** variables are now initialized **/
    *xPtr = c;
    *yPtr = d;

    printf("Next, lets get a point on the circle \nPlease enter the x and y coordinates, separated by a space: \n");
    scanf("%lf %lf", &c, &d);

    /** call distance function **/
    distance(a,b,c,d);
}
/** this function is gonna find this distance between two points **/
double distance(double x1, double x2, double y1, double y2) {

    /** initialize distance formula **/
    double formula1 = (x2-x1) * 2 * 2; 
    double formula2 = (y2-y1) * 2 * 2;
    double formula = formula1 + formula2;
    /** call sqrt function to square root it **/
    sqrt(formula);
    printf("\n%lf", sqrt(formula));

    return sqrt(formula);
}

我正在路过让我们说1 0＆amp; 0 0

Answer 1

以下内容未将(x2 - x1)提升至第二权力;相反，它乘以4：

double formula1 = (x2-x1) * 2 * 2;

下一行也是如此。

一旦你修复了这些，事情就会有所改善（你将不再尝试计算负数的实平方根，这就是给你NaN的原因。）

修改：要计算平方，请使用(x2-x1) * (x2-x1)或pow(x2-x1, 2.0)。

Answer 2

你将把你的距离函数称为

distance (x1=1, x2=0, y1=0, y2=0)

然后

double formula1 = (x2-x1) * 2 * 2;

将评估为-4

然后

 double formula = formula1 + formula2;

将评估为-4

从而

sqrt(formula);

as -nan

sqrt（）函数在c中不起作用我的数学错误是什么？

2 个答案: