skrtbhtngr已经回答了陈述的问题,但我想解决潜在的问题。
应用Unix philosophy。使用一个工具生成数据,另一个使用它 - 在这里,使用梯形法则计算积分。
您可以使用最简单的格式是使用Gnuplot支持的相同:
- 空行被忽略
- 线,
#
开始被忽略,可以用于注释
- 每一行定义一个样品
本质上讲,你可以描述一个正弦曲线非常粗略地使用
#x sin(x)
0.000 0.000000000
0.100 0.099833417
0.200 0.198669331
0.300 0.295520207
0.400 0.389418342
0.500 0.479425539
0.600 0.564642473
0.700 0.644217687
0.800 0.717356091
0.900 0.783326910
1.000 0.841470985
1.100 0.891207360
1.200 0.932039086
1.300 0.963558185
1.400 0.985449730
1.500 0.997494987
1.600 0.999573603
1.700 0.991664810
1.800 0.973847631
1.900 0.946300088
2.000 0.909297427
2.100 0.863209367
2.200 0.808496404
2.300 0.745705212
2.400 0.675463181
2.500 0.598472144
2.600 0.515501372
2.700 0.427379880
2.800 0.334988150
2.900 0.239249329
3.000 0.141120008
3.100 0.041580662
3.200 -0.058374143
3.300 -0.157745694
3.400 -0.255541102
3.500 -0.350783228
3.600 -0.442520443
3.700 -0.529836141
3.800 -0.611857891
3.900 -0.687766159
4.000 -0.756802495
4.100 -0.818277111
4.200 -0.871575772
4.300 -0.916165937
4.400 -0.951602074
4.500 -0.977530118
4.600 -0.993691004
4.700 -0.999923258
4.800 -0.996164609
4.900 -0.982452613
5.000 -0.958924275
5.100 -0.925814682
5.200 -0.883454656
5.300 -0.832267442
5.400 -0.772764488
5.500 -0.705540326
5.600 -0.631266638
5.700 -0.550685543
5.800 -0.464602179
5.900 -0.373876665
6.000 -0.279415498
6.100 -0.182162504
6.200 -0.083089403
你可以使用例如。 awk来产生,像我一样:
awk 'BEGIN { printf "#x sin(x)\n" ; for (x=0.0; x<6.3; x+=0.1) printf "%.3f %11.9f\n", x, sin(x) }'
如果您保存到一个文件中(追加> data.txt
上述命令),你可以在gnuplot的使用
plot "data.txt" using 1:2 notitle with lines
这些数据绘制很容易阅读C程序。因为我只使用POSIX.1系统(Linux,BSD和macOS),而POSIX.1提供了非常有用的getline()
函数 - 它可以让你读取任意长度的行,动态分配足够大的缓冲区 - 这个特定的实现还需要POSIX.1支持。换句话说,除了Windows以外,它基本上都可以工作。
#define _POSIX_C_SOURCE 200809L
#include <stdlib.h>
#include <string.h>
#include <stdio.h>
#include <errno.h>
/* Read x y(x) values, one pair per line, from a stream.
The arrays are dynamically allocated, and pointers stored
to *xptr and *yptr. The size of the arrays is stored at *nptr.
They are initially cleared to NULL/zero.
The function returns 0 if success, an errno error code otherwise:
EINVAL: Invalid function parameters
EIO: Read error
ENOMEM: Out of memory
EBADMSG: Malformed line
*/
int read_xy(double **xptr, double **yptr, size_t *nptr, FILE *in)
{
/* Line input buffer variables. */
char *line = NULL;
size_t size = 0;
ssize_t len;
/* Data array variables. */
double *x = NULL;
double *y = NULL;
size_t n = 0; /* Entries in x[] and y[] */
size_t nmax = 0; /* Entries allocated */
/* Temporary variables. */
double xval, yval, *newx, *newy;
/* We clear the output parameters to NULL or zero,
in case the caller is careless and does not check
the return value. Clearing them ensures they do
not contain garbage in such a case. */
if (xptr)
*xptr = NULL;
if (yptr)
*yptr = NULL;
if (nptr)
*nptr = 0;
/* We need in and nptr, and at least one of xptr and yptr. */
if (!in || !nptr || (!xptr && !yptr))
return errno = EINVAL;
/* If an error has already occurred in 'in',
we do not even try to read from it. */
if (ferror(in))
return EIO;
while (1) {
/* Read next input line. */
len = getline(&line, &size, in);
/* End of input or error? */
if (len < 0)
break;
/* Skip empty and comment lines. */
if (len == 0 ||
line[0] == '\n' || (line[0] == '\r' && line[1] == '\n') ||
line[0] == '#')
continue;
/* Parse the line. */
if (sscanf(line, " %lf %lf", &xval, &yval) != 2)
break;
/* Need to grow the dynamically allocated arrays? */
if (n >= nmax) {
/* Allocation policy.
We allocate room for at least 16 doubles,
then double the size up to 1048576 (=2^20),
then adjust to the next full multiple of 1048576.
This is not 'the best', but it is robust,
and not too wasteful.
*/
if (n < 16)
nmax = 16;
else
if (n < 1048576)
nmax = n * 2;
else
nmax = (n | 1048575) + 1048576;
/* Note: realloc(NULL, size) is equivalent to malloc(size).
If the realloc() call fails, it returns NULL,
but the original array is still valid.
Also note that free(NULL) is safe, and does nothing.
*/
newx = realloc(x, nmax * sizeof x[0]);
newy = realloc(y, nmax * sizeof y[0]);
if (newx)
x = newx;
if (newy)
y = newy;
if (!newx || !newy) {
/* One or both of the allocations failed. */
free(line);
free(x);
free(y);
return ENOMEM;
}
}
/* Save the parsed values to the arrays. */
x[n] = xval;
y[n] = yval;
n++;
}
/* We no longer need the line buffer. */
free(line);
/* Did a read error occur? */
if (ferror(in)) {
free(x);
free(y);
return EIO;
}
/* Was there no data to read? */
if (n < 1) {
free(x);
free(y);
return 0;
}
/* Reallocate the arrays to their exact sizes
(actually, allow for one extra double at the end,
because it is often useful to copy the initial
ones there if the data is considered cyclic).
*/
nmax = n + 1; /* One extra just because it is so often useful. */
newx = realloc(x, nmax * sizeof x[0]);
newy = realloc(y, nmax * sizeof y[0]);
if (newx)
x = newx;
if (newy)
y = newy;
if (!newx || !newy) {
free(x);
free(y);
return ENOMEM;
}
/* Save the array pointers. */
if (xptr)
*xptr = x;
else
free(x);
if (yptr)
*yptr = y;
else
free(y);
/* Save the number of samples read. */
*nptr = n;
/* If feof(in) is true, then we read everything
up to end of input. Otherwise, we stopped at
a line we could not parse.
*/
if (!feof(in))
return EBADMSG;
return 0;
}
该函数或类似的东西应该在每个数值计算课程的课程材料中。他们只是非常有用。除了系统管理员为每个进程设置的可能的内存分配限制以外,这个特定的内存对它可以读取的数据的大小没有固有的限制。我知道一个事实,即如果你只有足够的内存,它可以成功读取数十亿行数据。
使用该功能非常简单。以下是一个示例main()
,它只是从标准输入中读取这些数据 - 请记住,您可以通过在运行时将< file
附加到命令中来从文件中读取数据,并将数据输出。
int main(void)
{
double *x, *y;
size_t i, n;
int result;
result = read_xy(&x, &y, &n, stdin);
switch (result) {
case 0: /* No errors */
break;
case EBADMSG:
if (n > 1)
fprintf(stderr, "Invalid line after %zu data samples.\n", n);
else
fprintf(stderr, "Cannot parse first input line.\n");
return EXIT_FAILURE;
case ENOMEM:
fprintf(stderr, "Out of memory.\n");
return EXIT_FAILURE;
case EIO:
fprintf(stderr, "Read error.\n");
return EXIT_FAILURE;
case EINVAL:
fprintf(stderr, "Invalid parameters to the read_xy() function!\n");
return EXIT_FAILURE;
default:
fprintf(stderr, "%s.\n", strerror(result));
return EXIT_FAILURE;
}
printf("Read %zu samples:\n", n);
for (i = 0; i < n; i++)
printf("%.9f %.9f\n", x[i], y[i]);
return EXIT_SUCCESS;
}
注意,大部分是switch (result) { .. }
错误报告代码。这个例子非常小心的告诉你,如果发生错误,原因在于,作为用户,你需要知道什么时候知道他们可能会喷出垃圾,并且在现实生活中会倾向于程序放弃而不是静静地扔出垃圾 - 也许会让你相信它是有效的。
虽然可以修改上面的代码,甚至在Windows上运行(你getline()
与fgets()
取代,例如,希望你使用它的缓冲区大小就足够了;而且可能还需要改变一些错误代码errno
)。然而,在Top 500超级计算机列表中没有Windows机器是有原因的:POSIXy系统(Unix和Linux)更适合科学计算。所以,如果你打算开展一些科学计算,那么你可能只需要建立一个Linux或BSD虚拟机,然后在那里进行开发。
也许有另一个去格式化代码,并使其更具可读性。 PS什么_感谢你的优势_是什么意思? –
完成。重要的部分是功能“功能”。现在这个函数是硬编码的,我想把它作为一个参数来读取,例如从命令行。 – Noveel
听起来像你需要一个解析器。 – EOF