函数未提供有效输出（Visual Studio 2010上的C）

我曾尝试将我在电子方面所做的微积分自动化。不幸的是，我的代码似乎不起作用，因为所有结果都是0.000000（见下文）

以下是我尝试过的： -双精度或整型变量 -移除微积分部分并尝试从输入中获取输出 -扫描和扫描

这两个都不起作用。奇怪的是，最后的printf甚至不能打印base_v（例如）

下面是代码本身：

#include "stdafx.h"
#include "stdio.h"
#include "stdlib.h"
#include "math.h"

int _tmain(int argc, _TCHAR* argv[])
{
    // variable declaration
    double base_r, mod_r, base_v, mod_v=0;
    double coeff=1;
    int asdf=1;

    // welcome message
    printf("vMod calculator\n");
    printf("Please enter resistances in Ohms and voltages in Volts\n");

    // retrieve info about vmod
    printf("Enter base resistance\n");
    scanf_s("%f", &base_r);
    printf("Enter modified resistance\n");
    scanf_s("%f", &mod_r);
    printf("Enter base voltage\n");
    scanf_s("%f", &base_v);

    // calculus of modified voltage
    coeff=asdf*((1/base_r)+(1/mod_r));
    mod_v=base_v*base_r*coeff;


    // results
    printf("Base voltage: %f \n", &base_v);
    printf("Base resistance: %f \n", &base_r);
    printf("Modified voltage: %f \n", &mod_v);
    printf("Modified resistance: %f \n", &mod_r);

    system("pause");

    return 0;

}

如果您不理解微积分，请参考下图：

---

谢谢

您传递的是

printf

参数

&base\u v

的地址，而不是它的值

base\u v

。简而言之，摆脱

&

printf("Base voltage: %f \n", base_v);
printf("Base resistance: %f \n", base_r);
printf("Modified voltage: %f \n", mod_v);
printf("Modified resistance: %f \n", mod_r);

---

您传递的是

printf

参数

&base\u v

的地址，而不是它的值

base\u v

。简而言之，摆脱

&

printf("Base voltage: %f \n", base_v);
printf("Base resistance: %f \n", base_r);
printf("Modified voltage: %f \n", mod_v);
printf("Modified resistance: %f \n", mod_r);

---

首先，您将指针打印为浮点数，因此具有未定义的行为。

首先，您将指针打印为浮点数，因此具有未定义的行为。

printf

直接获取变量，而不是其地址：

printf("Base voltage: %f \n", base_v);
printf("Base resistance: %f \n", base_r);
printf("Modified voltage: %f \n", mod_v);
printf("Modified resistance: %f \n", mod_r);

---

printf

直接获取变量，而不是其地址：

printf("Base voltage: %f \n", base_v);
printf("Base resistance: %f \n", base_r);
printf("Modified voltage: %f \n", mod_v);
printf("Modified resistance: %f \n", mod_r);

---

scanf在第一次调用后将换行符0x13和0x10保留在stdin缓冲区中，所以对于dos/windows，下一次调用将首先获得13/10 使用fgets，然后使用fscanf或类似的东西 而（（dumchar=getch（））==\n）；

---

要在每次scanf（）调用之前清除换行符，scanf会在第一次调用后在stdin缓冲区中保留换行符0x13和0x10，因此对于dos/windows，下一次调用将首先获得13/10 使用fgets，然后使用fscanf或类似的东西 而（（dumchar=getch（））==\n）；

---

为了在每次scanf（）调用之前清除换行符

我尝试不使用

&

，现在我得到了四个“-92559604647977569000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000。也许您可以在

scanf_s

调用中使用float或

%lf

来尝试这一点。谢谢我尝试不使用

&

，现在我得到了四个“-92559604647977569000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000。也许您可以在

scanf_s

调用中使用float或

%lf

来尝试这一点。谢谢