gpt4 book ai didi

c - 功能无法按照我希望的那样工作 (c)

转载 作者:行者123 更新时间:2023-11-30 21:31:06 26 4
gpt4 key购买 nike

我必须编写一个能够读取 .txt 文件的程序。在此 .txt 文件中是传感器上激光束的测量结果,传感器前面有 Blade 阻挡光束。慢慢地, Blade 下降并测量光强度。我必须编写一个程序,用 txt 文件中的测量值计算激光束的宽度(给出了公式,(((0,9*max)-(0,1*max))/1,28 )。

但是如果我在没有函数的情况下执行此操作,它就可以正常工作。但这个程序有一些要求。我必须使用函数。你可能会认为我的程序有点狡猾,但这主要是因为我必须这样做。如果有些事情令人困惑,请向我询问更多解释,我会立即给您。

这是我当前的程序:

#include <stdio.h>
#define ROW 45
#define COLUMN 2

FILE *measurements;
float x,width, intensity, a[ROW][COLUMN]={0}, total_2 = 0, maximum = 0;
float max = 0, min = 0, difference_1 = 0, difference_2 = 0, w = 0;
float background=0, amount=0, total_1=0;
int menu=0, check_1 = 0, check_2 = 0, min_2 = 0, max_2 = 0;

//Functions
float background_radiation(float a[ROW][COLUMN]);
float average_maximum(float a[ROW][COLUMN]);
float beam_width(float a[ROW][COLUMN], float maximum);

//Integers for loops
int i = 0, r = 0, k = 0, d = 0;

main()
{
measurements = fopen("PATH\\TO\\TEXT\\FILE\\.txt", "r");
rewind(measurements);

//Menu
printf("Menu: \n\n");
printf("1. Calculate background_radiation: \n");
printf("2. Calculate average maximum signal: \n");
printf("3. Calculate beam width: \n");
printf("4. All measurements: \n");
printf("5. Quit \n\n");
printf("\nChoose:\n");

scanf("%d", &menu);
rewind(measurements);

//All data in array
while(!feof(measurements))
{
for(i=0; i<45; i++)
{
fscanf(measurements, "%f%f", &a[i][0], &a[i][1]);
}
}

//Menu loop
while (menu!=5)
{
switch(menu)
{
case 1: printf("\n------ background_radiation ------ \n");
//case check
check_1 = 1;

x= background_radiation(a);
for(i=0; i<45; i=i+1)
{
a[i][1]=a[i][1]-x;
}
break;

case 2: printf("\n------ Maximale signaal ------ \n");
//case check
check_2 = 1;

y = average_maximum(a);

break;

case 3: printf("\n------ beam_width ------ \n");
//check if the cases 1 and 2 are used
if(check_1 == 1 && check_2 == 1)
{
z = beam_width(a, y);
}

else
{
//If case 1 and 2 aren't used
printf("The backgroundradiation and maximum output aren't calculated\n");

printf("You have 2 options:\n");
printf("1. Use the first and last measurements to calculate the beamwidth.\n");
printf("2. Choose an other option of the menu.\n");
scanf("%d", &d);
//Print the beam width
if(d==1)
{
printf("\nBeamwidth:\t%f mm", ((a[0][44]-a[0][0])/1.28));
}
}
break;

case 4: printf("\n------------ Measurements ------------\n\n");
rewind(measurements);
while(!feof(measurements))
{
fscanf(measurements, "%f%f", &a[i][0], &a[i][1]);
printf("\t%2.2f mm\t\t%2.2f V\n", a[i][0], a[i][1]);
}
break;
}
rewind(measurements);
printf("\n\nChoose an option: ");
scanf("%d", &menu);
}

fclose(measurements);
return 0;
fflush(stdin);

}

float background_radiation(float a[ROW][COLUMN])
{
printf("\nHow many numbers would you like to take an average?\n");
scanf("%d", &amount);

for(i=0; i<amount; i++)
{
total_1 = total_1 + a[i][1];
}

//Calculate average
background = total_1 / amount;
printf("\nThe average background_radiation:\n%2.2f\n", background);

return(background);
}

float average_maximum(float a[ROW][COLUMN])
{
printf("\nHow many numbers would you like to take an average?\n");
scanf("%d", &amount);

for(i=44; i>(44-amount); i--)
{
total_2 = total_2 + a[i][1];
}

//Calculate average
maximum = total_2 / amount;
printf("\nAverage maxixum signal:\n%2.2f\n", maximum);
return(maximum);
}

float beam_width(float a[ROW][COLUMN], float y)
{
max = 0.9 * y;
min = 0.1 * y;

//Find a point in array
for(r=0; r<45; r++)
{
difference_1 = max - a[r][1];
if ((difference_1 < difference_2) && difference_1 > 0)
{
max_2 = r;
}
difference_2 = difference_1;
}

difference_2 = 100;

for(k=0; k<45; k++)
{
difference_1 = min - a[k][1];
if (difference_1 < difference_2 && difference_1>0)
{
min_2 = k;
}
difference_2 = difference_1;
}

//calculate width with the given formula
w = (a[max_2][0] - a[min_2][0]) / 1.28;
printf("\nBeamwidth:\t%2.2f", w);
return(w);
}

文本文件由两列和 45 行组成。第一列是 Blade 向下的高度。所以如果它是0。这意味着 Blade 完全挡住了光束。第二列是光强度。由于传感器周围的光(背景辐射),它始终具有一个值。这是文本文件:

0.00 0.25
0.10 0.20
0.20 0.18
0.30 0.21
0.40 0.23
0.50 0.30
0.60 0.30
0.70 0.40
0.80 0.50
0.90 0.80
1.00 1.30
1.10 1.80
1.20 2.30
1.30 3.80
1.40 4.50
1.50 6.30
1.60 8.04
1.70 10.55
1.80 13.10
1.90 16.20
2.00 19.80
2.10 22.56
2.20 25.10
2.30 29.90
2.40 31.20
2.50 33.44
2.60 36.80
2.70 41.05
2.80 40.83
2.90 43.40
3.00 44.44
3.10 44.90
3.20 45.40
3.30 46.00
3.40 46.30
3.50 46.50
3.60 46.60
3.70 46.50
3.80 46.35
3.90 46.40
4.00 46.60
4.10 46.30
4.20 46.00
4.30 45.90
4.40 46.00

您可能会认为第二列在开始时下降很奇怪,但这与测量的准确性有关。

非常感谢您!

什么不起作用?:

所有功能(background_radiation、average_maximum、beam_width)都不起作用。

情况 1 的输出(总是):
1.#J

案例 2 的输出(始终):
0.00

案例3的输出
如果使用情况 1 和 2:
0.00

其他
工作正常

最佳答案

yz 未定义时,此代码不会编译。在这些变量前面添加 float 并更改运行路径,我发现以下错误:

首先,将数据加载到 a 对我来说失败了:

a[ROW][COLUMN] = {0}

产生了一个垃圾数组,所以我放弃了初始化以获得:

a[ROW][COLUMN]

允许数据加载工作。

下一个问题是您希望将 amount 视为 float 和整数。在background_radiation()中,我为循环创建了一个额外的参数并更改了scanf():

scanf("%f", &amount);

int max = (int)amount;
for(i=0; i<max; i++)

在这些更改之后,如果我输入“12.0”作为 amount float ,我会得到以下输出:

------ background_radiation ------ 

How many numbers would you like to take an average?
12.0

The average background_radiation:
0.54

average_maximum() 可以类似地工作。 beam_width() 可能遇到数据加载问题。它似乎对我来说工作得很好。

关于c - 功能无法按照我希望的那样工作 (c),我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/19401304/

26 4 0
Copyright 2021 - 2024 cfsdn All Rights Reserved 蜀ICP备2022000587号
广告合作:1813099741@qq.com 6ren.com