[英]What is the fastest way to count lines and words in a text file in ANSI C?
What is the fastest way to count lines and words in a text file in pure ANSI C? 用纯ANSI C计算文本文件中行和单词的最快方法是什么?
A word is terminated by a space or period. 单词以空格或句号结尾。 Line is terminated by '\\n'
. 行以'\\n'
终止。
Maybe take a look at the source code of the GNU wc utility as this utility does exactly what you want. 也许请看一下GNU wc实用程序的源代码,因为该实用程序确实可以满足您的需求。
#include <stdlib.h>
#include <stdio.h>
#include <stdarg.h>
typedef unsigned long count_t; /* Counter type */
/* Current file counters: chars, words, lines */
count_t ccount;
count_t wcount;
count_t lcount;
/* Totals counters: chars, words, lines */
count_t total_ccount = 0;
count_t total_wcount = 0;
count_t total_lcount = 0;
/* Print error message and exit with error status. If PERR is not 0,
display current errno status. */
static void
error_print (int perr, char *fmt, va_list ap)
{
vfprintf (stderr, fmt, ap);
if (perr)
perror (" ");
else
fprintf (stderr, "\n");
exit (1);
}
/* Print error message and exit with error status. */
static void
errf (char *fmt, ...)
{
va_list ap;
va_start (ap, fmt);
error_print (0, fmt, ap);
va_end (ap);
}
/* Print error message followed by errno status and exit
with error code. */
static void
perrf (char *fmt, ...)
{
va_list ap;
va_start (ap, fmt);
error_print (1, fmt, ap);
va_end (ap);
}
/* Output counters for given file */
void
report (char *file, count_t ccount, count_t wcount, count_t lcount)
{
printf ("%6lu %6lu %6lu %s\n", lcount, wcount, ccount, file);
}
/* Return true if C is a valid word constituent */
static int
isword (unsigned char c)
{
return isalpha (c);
}
/* Increase character and, if necessary, line counters */
#define COUNT(c) \
ccount++; \
if ((c) == '\n') \
lcount++;
/* Get next word from the input stream. Return 0 on end
of file or error condition. Return 1 otherwise. */
int
getword (FILE *fp)
{
int c;
int word = 0;
if (feof (fp))
return 0;
while ((c = getc (fp)) != EOF)
{
if (isword (c))
{
wcount++;
break;
}
COUNT (c);
}
for (; c != EOF; c = getc (fp))
{
COUNT (c);
if (!isword (c))
break;
}
return c != EOF;
}
/* Process file FILE. */
void
counter (char *file)
{
FILE *fp = fopen (file, "r");
if (!fp)
perrf ("cannot open file `%s'", file);
ccount = wcount = lcount = 0;
while (getword (fp))
;
fclose (fp);
report (file, ccount, wcount, lcount);
total_ccount += ccount;
total_wcount += wcount;
total_lcount += lcount;
}
int
main (int argc, char **argv)
{
int i;
if (argc < 2)
errf ("usage: wc FILE [FILE...]");
for (i = 1; i < argc; i++)
counter (argv[i]);
if (argc > 2)
report ("total", total_ccount, total_wcount, total_lcount);
return 0;
}
Found at: http://www.gnu.org/software/cflow/manual/html_node/Source-of-wc-command.html 可在以下网址找到: http : //www.gnu.org/software/cflow/manual/html_node/Source-of-wc-command.html
EOF
重复第二和第三步,直到EOF
Here is an explicit answer that counts the number of lines (extension to the number of words is trivial à la the C++ version linked to in OP). 这是一个对行数进行计数的显式答案(对单词数的扩展对于在OP中链接到的C ++版本来说是微不足道的)。 This version is buffered. 此版本已缓冲。 Another answer suggests reading the entire file in first, which is simpler, but the below is more in line with what your C++ example does. 另一个答案是建议首先读取整个文件,这比较简单,但是下面的内容与您的C ++示例所做的更加一致。
#include <stdio.h>
#include <string.h>
#define BUFSIZE 1024
int main(int argc, char** argv)
{
int newlines = 0;
char buf[BUFSIZE];
FILE* file;
if (argc != 2)
return 1;
file = fopen(argv[1], "r");
while (fgets(buf, BUFSIZE, file))
{
if (!(strlen(buf) == BUFSIZE-1 && buf[BUFSIZE-2] != '\n'))
newlines++;
}
printf("Number of lines in %s: %d\n", argv[1], newlines);
return 0;
}
The BUFSIZE macro can be tweaked to maximize performance (since you say you want the fastest way). 可以调整BUFSIZE宏以最大化性能(因为您说您想要最快的方法)。 1024 is simply a guess. 1024只是一个猜测。 Another possibility is probably to read the file memory mapped, but I didn't try since mmap is not ANSI C. 另一种可能性可能是读取映射的文件内存,但是我没有尝试,因为mmap不是ANSIC。
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