Question

此容差分析代码编译并运行，但输出不正确，我认为这是对文件和文件中数据的错误处理。

文件中的数据是：

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

代码：

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

void parsePart(char input[], float*pnom, int*pinp, float*ptol, char*pFV);
void parseGap(char input[], float *pmin, float *pmax);
float meanGap(float nom[], float imp[], int size);
float tolGap(float tol[], int size);

int main()
{
    FILE *ptable;
    int i, num_of_parts;
    int impact[10];
    float nominal[10], tolerance[10];
    char FV[10];
    char input_str[30];
    float max, min, gap, tol;
    float curr_max, curr_min;

    ptable=fopen("D:\\Input.txt", "r");

    for(i=0; i<11; i++);
    {
        fgets(input_str, 30, ptable);
        if(input_str[0] == 'P')
        {
            parsePart(input_str, &nominal[i], &impact[i], &tolerance[i], &FV[i]);
        }
        else
        {
            parseGap(input_str, &min, &max);
            num_of_parts = i;

        }
    }

    gap = meanGap(nominal, impact, num_of_parts);
    tol = tolGap(tolerance, num_of_parts);
    curr_max = gap+tol;
    curr_min = gap-tol;

    printf("Gap mean is: %f inches\n", gap);
    printf("Gap tolerance is: %f inches\n", tol);

    if(fabs(max-curr_max)< 0.0001 && curr_max > (gap*2)) //they are equal
    {
        printf("The maximum gap is %f which is greater than the allowed %f\n", curr_max, gap*2);
    }
    else
    {
        printf("The maximum gap is %f which is within the allowed %f\n", curr_max, gap*2);
    }
    if(fabs(min+curr_min)<0.0001 && curr_min < gap-gap)
    {
        printf("The minimum gap is %f which is less than the allowed %f\n", curr_min, gap-gap);
    }
    else
    {
        printf("The minimum gap is %f which is within the allowed %f\n", curr_min, gap-gap);
    }
    return 0;
}

void parsePart(char input[], float*pnom, int*pinp, float*ptol, char*pFV)
{
    int i;
    char * field[5];

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

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

    *pnom = atof(field[1]);
    *pinp = atoi(field[2]);
    *ptol = atof(field[3]);
    *pFV = *field[4];
}

void parseGap(char input[], float *pmin, float *pmax)
{
    char *field[2];

    field[0] = strtok(input, ",");
    field[1] = strtok(NULL, ",");
    field[2] = strtok(NULL, ",");

    *pmin = atof(field[1]);
    *pmax = atof(field[2]);
}

float meanGap(float nom[], float imp[], int size)
{
    int i;
    float sum=0;

    for(i=0; i<size; i++);
    {
        sum += nom[i]*imp[i];
    }
    return sum;
}
float tolGap(float tol[], int size)
{

    int i;
    float sum=0;

    for(i=0; i<size; i++);
    {
        sum += tol[i];
    }
    return sum;
}

输出应该类似于：

Actual Gap Mean: 0.025”
Actual Gap Tolerance: 0.085”
The Maximum Gap (0.110”) is (Greater) than specified (0.080”)
The Minimum Gap (-0.060”) is (Less) than the specified (0.000”)

我得到0表示差距，而所有其他值都是非常大的数字。

关于我可能出错或需要改进的地方的任何和所有提示都很棒。谢谢。

Answer 1

在第85行，char *field[2];应为char *field[3];。

Answer 2

首先，数组影响等的大小为10，所以

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

错误有两个原因：

1）循环应该从0到9

2） for 语句不应该有;在它之后，它终止了整个陈述。

Answer 3

  typedef struct{char c[6];} String;
  String field1,field2,… …;
  String *pf1,*pf2,… …;

  pf1=field1;
  pf2=field2;
  …
  …
  pf1=strtok(input,…

是否正确（大纲）代码 parseGap ，假设输入文件中的最长数据长度为5个字符，

FILE I / O问题并使用

3 个答案: