C++ 声明、定义和调用

我需要更好地理解使用这个程序的函数定义、声明和正确调用。我真的需要了解如何使用它们。你能告诉我正确的方法来编写这个程序吗？这三种方法都是正确的，并且解释得很清楚

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

quad_equation(float a, float b, float c);

int main()

{

    float a, b, c, determinant, r1,r2;

    printf("Enter coefficients a, b and c: ");

    scanf("%f%f%f",&a,&b,&c);

    determinant=b*b-4*a*c;

    if (determinant>0)

    {

        r1= (-b+sqrt(determinant))/(2*a);

        r2= (-b-sqrt(determinant))/(2*a);

        printf("Roots are: %.2f and %.2f",r1 , r2);

    }

    else if (determinant==0) { r1 = r2 = -b/(2*a);

    printf("Roots are: %.2f and %.2f", r1, r2);

    }


    else (determinant<0);

    {

    printf("Both roots are complex");

    }

    return 0;

#包括
#包括
四元方程（浮点a、浮点b、浮点c）；
int main（）
{
浮点a，b，c，行列式，r1，r2；
printf（“输入系数a、b和c:”）；
scanf（“%f%f%f”、&a、&b和&c）；
行列式=b*b-4*a*c；
如果（行列式>0）
{
r1=（-b+sqrt（行列式））/（2*a）；
r2=（-b-sqrt（行列式））/（2*a）；
printf（“根是：%.2f和%.2f”，r1，r2）；
}
如果（行列式==0）{r1=r2=-b/（2*a）；
printf（“根是：%.2f和%.2f”，r1，r2）；
}
else（行列式我刚刚解决了这个问题：（我想这是作业的一部分）


同时看看你的代码。你从来没有使用过函数四元方程。而且你还没有定义函数的类型（int/void/float/char）等等

为方便起见：（这是完整的代码）--如果你什么都不懂，请问我

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

// function declarations

void twoRoots (float a,float b,float delta);
void oneRoot (float a,float b,float delta);

 int main (void)                        
    {
    //Local Declarations
    float a;
    float b;
    float c;
    float delta;

   // float solution;

    printf("Input coefficient a.\n");
    scanf("%f", &a);
    printf("Input coefficient b.\n");
    scanf("%f", &b);
    printf("Input coefficient c.\n");
    scanf("%f", &c);
    printf("%0.2fx^2 + %0.2fx + %0.2f\n", a, b, c);

    delta = (float)(b*b) - (float)(4.0 * a * c);

    printf("delta = %0.2f\n",delta);

    if (delta > 0){
         twoRoots(a,b,delta);
    }else if (delta == 0) {
         oneRoot(a,b,delta);
    }else if (delta < 0.0){
         printf("There are no real roots\n");
    }

    return 0;
} 

void twoRoots (float a,float b,float delta)
{

    float xOne;
    float xTwo;

    float deltaRoot;

    printf("There are two distinct roots.\n");
    deltaRoot = sqrt(delta);
    xOne = (-b + deltaRoot) / (2*a);
    xTwo = (-b - deltaRoot) / (2*a);
    printf("%.2f", xOne);
    printf("%.2f", xTwo);
} 



void oneRoot(float a,float b,float delta)
{

    float xOne;
  //  float xTwo;
   // float deltaRoot;

    printf("There is exactly one distinct root\n");
    xOne = -b / (2*a);
    printf("%.2f", xOne);

}

如果您输入以下内容，此操作将失败：
121

因为scanf将把整个
121
读入
a
，而
b
，
c
将没有任何内容（而是将\n（输入）放入b，未定义到c）

因此，请按照我在代码中使用scanf的方式使用scanf，好吧-这充满了问题！我试图指出这些问题，并展示“更好”是什么样子。我希望这会有所帮助

quad_equation(float a, float b, float c);

这可能是一个“函数原型”。原型告诉编译器“我稍后将使用这个函数，这就是调用它的方式，以及它返回的类型”。您没有指定返回类型；您可能希望使用
int
说明是否找到了根，并在函数中打印结果。最好将两个返回值的空格作为参数传递：

int quad_equation(float a, float b, float c, float* x1, float* x2);

现在我们可以使用
main
程序获取输入/输出，让函数解决问题：

int main(void) {
{
float a, b, c, r1, r2;
int n;

// here you get the inputs; that seems OK
printf("Enter coefficients a, b and c: ");
scanf("%f %f %f",&a,&b,&c);

// now you have to "call your function"
// note that I make sure to follow the prototype: I assign the return value to an int
// and I pass five parameters: the coefficients a, b, c and the address of two variables
// x1 and x2. These addresses will be where the function puts the roots
n = quad_equation(a, b, c, &r1, &r2);

// when the function returns, I can print the results:
printf("There are %d roots:\n", n);

// based on the value of n, I change what I want to print out:
if (n == 2) printf(" %f and ", r1);  // when there are two roots I print "root 1 and"
if (n > 0) printf("%f\n", r2);       // when there is at least one root, I print it
// note that if n == 0, I would skip both print statements
// and all you would have gotten was "There are 0 roots" in the output

}

int quad_equation(float a, float b, float c, float* x1, float* x2) {
// function that computes roots of quadratic equation
// and returns result in x1 and x2
// it returns the number of roots as the return value of the function

float determinant;

  determinant=b*b-4*a*c;

  if (determinant>0)
  {
    *x1 = (-b+sqrt(determinant))/(2*a);
    *x2= (-b-sqrt(determinant))/(2*a);
    return 2;
  }

if (determinant==0) { 
  *x1 = *x2 = -b/(2*a);
  return 1;
}

return 0;
}

在声明中，我想知道“*”是什么意思浮点数旁边是表示？R1和R2是否也被假定为底部的X1和X2？*表示这不是一个
浮点数
，而是一个指向浮点数的指针-也就是说，这个变量包含一个存储浮点数的位置的地址。这允许函数修改该位置的内容-因此它可以传递更多信息我故意在函数中给他们一个不同的名称，因为这是你可以做的事情之一：你用来定义它的变量的名称不需要是你调用它的变量的名称。第一个浮点解决方案“未使用的变量声明”，最底层的xtwo和deltar也是如此oot@user2879277删除它们..好的，等等..我会编辑代码是的，这就是我所做的，我重新安排了一些事情，因为这与我最初所做的大不相同。我真的很感谢你的帮助。如果你什么都不懂，问我..如果这有助于你，请接受答案是的，没问题，但出于某种原因，我拒绝了i don’我输入的每一件东西都没有真正的根，我想不出来
int main(void) {
{
float a, b, c, r1, r2;
int n;

// here you get the inputs; that seems OK
printf("Enter coefficients a, b and c: ");
scanf("%f %f %f",&a,&b,&c);

// now you have to "call your function"
// note that I make sure to follow the prototype: I assign the return value to an int
// and I pass five parameters: the coefficients a, b, c and the address of two variables
// x1 and x2. These addresses will be where the function puts the roots
n = quad_equation(a, b, c, &r1, &r2);

// when the function returns, I can print the results:
printf("There are %d roots:\n", n);

// based on the value of n, I change what I want to print out:
if (n == 2) printf(" %f and ", r1);  // when there are two roots I print "root 1 and"
if (n > 0) printf("%f\n", r2);       // when there is at least one root, I print it
// note that if n == 0, I would skip both print statements
// and all you would have gotten was "There are 0 roots" in the output

}

int quad_equation(float a, float b, float c, float* x1, float* x2) {
// function that computes roots of quadratic equation
// and returns result in x1 and x2
// it returns the number of roots as the return value of the function

float determinant;

  determinant=b*b-4*a*c;

  if (determinant>0)
  {
    *x1 = (-b+sqrt(determinant))/(2*a);
    *x2= (-b-sqrt(determinant))/(2*a);
    return 2;
  }

if (determinant==0) { 
  *x1 = *x2 = -b/(2*a);
  return 1;
}

return 0;
}