我将如何设置函数来计算圆的面积 - C
How would I set up function to calculate the area of a circle - C
我有这个 C 程序需要计算用户输入的圆的面积。
我必须在主要功能之前使用它:
void area_circum(double radius, double *area, double *circum);
我无法让程序运行。我正在使用主要功能和另一个名为 area_circum();
这是我现在拥有的:
#include <stdio.h>
void area_circum(double radius, double *area, double *circum);
int main(void) {
double radius, area, circum;
printf("Please enter the radius of the circle: ");
scanf("%f", &radius);
area_circum(radius, area, circum);
printf("Area of circle : %0.4f\n", area);
return 0;
}
void area_circum(double radius, double *area, double *circum) {
double PIE = 3.141;
double areaC = 0;
areaC = PIE * radius * radius;
}
当我构建它并 运行 它时,它起作用了,我输入了一个数字,但随后它 returns 0.00
#include <stdio.h>
void area_circum(double radius);
int main() {
double radius;
printf("Please enter the radius of the circle: ");
scanf("%lf", &radius);
area_circum(radius);
return 0;
}
void area_circum(double radius) {
double PIE = 3.141;
double areaC = 0;
areaC = PIE * radius * radius;
printf("Area of circle : %0.4f\n", areaC);
}
你快到了。您需要做的就是 return 计算出的值:
void area_circum(double radius, double *area) {
double PIE = 3.141;
*area = PIE * radius * radius;
}
并像这样打电话:
area_circum(radius, &area);
从函数的名称和签名可以看出,它应该同时计算面积和周长。同样很明显,您应该将两个 int 变量(或两个指向 int 的指针)的地址传递给它,然后将给出计算结果:
#include <stdio.h>
void area_circum(double radius, double *area, double *circum);
int main(void)
{
double radius, area, circum;
printf("Please enter the radius of the circle: ");
scanf("%lf", &radius); // Note: You should add a check to see if
// scanf() failed.
area_circum(radius, &area, &circum); // &area is the address of area
// &circum is the address of circum
printf("Area of circle : %0.4f\n", area);
printf("Circumference of circle : %0.4f\n", circum);
return 0;
}
void area_circum(double radius, double *area, double *circum)
{
double PIE = 3.1416;
*area = PIE * radius * radius;
*circum = PIE * radius * 2;
}
我有这个 C 程序需要计算用户输入的圆的面积。 我必须在主要功能之前使用它:
void area_circum(double radius, double *area, double *circum);
我无法让程序运行。我正在使用主要功能和另一个名为 area_circum();
这是我现在拥有的:
#include <stdio.h>
void area_circum(double radius, double *area, double *circum);
int main(void) {
double radius, area, circum;
printf("Please enter the radius of the circle: ");
scanf("%f", &radius);
area_circum(radius, area, circum);
printf("Area of circle : %0.4f\n", area);
return 0;
}
void area_circum(double radius, double *area, double *circum) {
double PIE = 3.141;
double areaC = 0;
areaC = PIE * radius * radius;
}
当我构建它并 运行 它时,它起作用了,我输入了一个数字,但随后它 returns 0.00
#include <stdio.h>
void area_circum(double radius);
int main() {
double radius;
printf("Please enter the radius of the circle: ");
scanf("%lf", &radius);
area_circum(radius);
return 0;
}
void area_circum(double radius) {
double PIE = 3.141;
double areaC = 0;
areaC = PIE * radius * radius;
printf("Area of circle : %0.4f\n", areaC);
}
你快到了。您需要做的就是 return 计算出的值:
void area_circum(double radius, double *area) {
double PIE = 3.141;
*area = PIE * radius * radius;
}
并像这样打电话:
area_circum(radius, &area);
从函数的名称和签名可以看出,它应该同时计算面积和周长。同样很明显,您应该将两个 int 变量(或两个指向 int 的指针)的地址传递给它,然后将给出计算结果:
#include <stdio.h>
void area_circum(double radius, double *area, double *circum);
int main(void)
{
double radius, area, circum;
printf("Please enter the radius of the circle: ");
scanf("%lf", &radius); // Note: You should add a check to see if
// scanf() failed.
area_circum(radius, &area, &circum); // &area is the address of area
// &circum is the address of circum
printf("Area of circle : %0.4f\n", area);
printf("Circumference of circle : %0.4f\n", circum);
return 0;
}
void area_circum(double radius, double *area, double *circum)
{
double PIE = 3.1416;
*area = PIE * radius * radius;
*circum = PIE * radius * 2;
}