c - 代码的输出全为零

Question

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

float tolGap(float Tol[], int size);
float meanGap(float Nom[], float Imp[], int size);
void parsePart(char input[], float *pNom, int *pImp, float *pTol, char* pFoV);
void parseGap(char input[], float *max, float *min);

int main()
{
    FILE *pIN;
    int i;
    int Impact[10];
    float Nominal[10], Tolerance[10];
    char FoV[10];
    char input_str[30];
    float max, min, tol;
    int Num_of_Parts;
    float curr_max, curr_min, mean;

    pIN=fopen("testingcode.txt", "r");

    for(i=0; i<11; i++)
    {
        fgets(input_str,30, pIN);
        if(input_str[0] == 'p')
        {
            parsePart(input_str, &Nominal[i], &Impact[i], &Tolerance[i], &FoV[i]);
        }
        else
        {
            parseGap(input_str, &min, &max);
            Num_of_Parts = i;
            break;
        }
    }

    mean = meanGap(Nominal, Impact, Num_of_Parts);
    tol = tolGap(Tolerance, Num_of_Parts);
    curr_max = mean+tol;
    curr_min = mean - tol;

    if (fabs(max - curr_max) < 0.00001) //they are equal
    {
        printf("The Maximum Gap (%fin) is equal to specified (%fin)\n", max, curr_max);
    }

    if ((max - curr_max) > 0.00001)
    {
        printf("The Maximum Gap (%fin) is Greater than specified (%fin)\n", max, curr_max);
    }
    else
    {
        printf("The Maximum Gap (%fin) is within specified (%fin)\n", max, curr_max);
    }

    if (fabs(min - curr_min) < 0.00001) //they are equal
    {
        printf("The Minimum Gap (%fin) is equal to specified (%fin)\n", min, curr_min);
    }

    if ((min - curr_min) > 0.00001)
    {
        printf("The Minimum Gap (%fin) is Greater than specified (%fin)\n", min, curr_min);
    }
    else
    {
        printf("The Minimum Gap (%fin) is Less than specified (%fin)\n", min, curr_min);
    }

    return 0;
}

float tolGap(float Tol[], int size)
{
    int i;
    float sum=0;

    for (i=0; i<size; i++)
    {
        sum+=Tol[i];
    }
    printf("Actual Gap Tolerance: %fin\n", sum);
    return sum;
}

float meanGap(float Nom[], float Imp[], int size)
{
    int i;
    float sum=0;

    for(i=0; i<size; i++)
    {
        sum+= Nom[i]*Imp[i];
    }
    printf("Actual Gap Mean: %fin\n", sum);
    return sum;
}

void parsePart(char input[], float *pNom, int *pImp, float *pTol, char* pFoV)
{
    int i;
    char *elements[5];

    elements[0]=strtok(input, ",");

    for(i=1;i<5;i++)
    {
        elements[i] = strtok(NULL, ",");
    }

    *pNom = atof(elements[1]);
    *pImp = atoi(elements[2]);
    *pTol = atof(elements[3]);
    *pFoV = *elements[4];

}

void parseGap(char input[], float *max, float *min)
{
    int i;
    char *elements[5];

    elements[0]=strtok(input, ",");

    for(i=1;i<5;i++)
    {
        elements[i] = strtok(NULL, ",");
    }

    *max = atof(elements[1]);
    *min = atof(elements[2]);
}

我的文件中包含以下信息:

PART,2.000,-1,0.050,V

PART,0.975,-1,0.025,V

PART,3.000,+1,0.010,F

GAP,0.000,0.080

文件中没有空行，为了清楚起见，我添加了它们。问题是我的计算都没有完成。我的所有输出都是 0.000000。我想让它实际计算一下，但找不到问题。任何帮助将不胜感激!

Answer 1

您正在传递一个 int 数组

int Impact[10];

...

mean = meanGap(Nominal, Impact, Num_of_Parts);

到一个需要float数组的函数

float meanGap(float Nom[], float Imp[], int size)

<小时/>

您在数组范围之外写入:

int Impact[10];
float Nominal[10], Tolerance[10];
char FoV[10];
...

for(i=0; i<11; i++)

应该是

for(i=0; i<10; i++)

<小时/>

正如 @user2340048 所指出的，PART 是大写的，您正在 if(input_str[0] == 'p' 中检查 p )