读一个字,写的时候,我们找到足够长字符串或必须:
#include <ctype.h>
#include <stdio.h>
#include <stdlib.h>
int
main(int argc, char *argv[])
{
size_t min_str_len = 4;
size_t buf_len = 4; /* Must greater than or equal to min_str_len */
char buf[buf_len], ch;
size_t out_len, last_len;
last_len = out_len = 0;
while (fread(&ch, 1, 1, stdin) > 0) {
if (isprint(ch)) {
buf[out_len++] = ch;
if (out_len >= buf_len) {
fwrite(buf, 1, out_len, stdout);
last_len += out_len;
out_len = 0;
}
}
else {
if (out_len + last_len >= min_str_len) {
fwrite(buf, 1, out_len, stdout);
#ifdef NEWLINE
fwrite("\n", 1, 1, stdout);
#endif
last_len = out_len = 0;
}
else {
out_len = 0;
}
}
}
exit(EXIT_SUCCESS);
}
如果你想每次读取多个字节,这种“至少4个连续打印字符“将使它有点棘手:
#include <assert.h>
#include <ctype.h>
#include <stdio.h>
#include <stdlib.h>
int
main(int argc, char *argv[])
{
size_t str_min_len = 4;
size_t buf_len = 1024; /* Must greater than or equal to str_min_len */
char in_buf[buf_len], out_buf[buf_len];
size_t out_len, in_len, last_len;
last_len = out_len = 0;
while ((in_len = fread(in_buf, 1, buf_len, stdin)) > 0) {
assert(out_len == 0);
for (size_t i = 0; i < in_len; i++) {
char ch = in_buf[i];
if (isprint(ch)) {
out_buf[out_len++] = ch;
}
else {
if (out_len + last_len >= str_min_len) {
fwrite(out_buf, 1, out_len, stdout);
#ifdef NEWLINE
/* Write a newline between strings. */
fwrite("\n", 1, 1, stdout);
#endif
last_len = 0;
}
out_len = 0;
}
}
if (0 < out_len && out_len < str_min_len) {
size_t pad_len = str_min_len - out_len;
for (size_t i = 0; i < pad_len; i++) {
char ch;
if (fread(&ch, 1, 1, stdin) < 1) {
exit(EXIT_SUCCESS);
}
else if (isprint(ch)) {
out_buf[out_len++] = ch;
}
else {
break;
}
}
}
if (out_len >= str_min_len) {
fwrite(out_buf, 1, out_len, stdout);
last_len = out_len;
out_len = 0;
}
else {
last_len = out_len = 0;
}
}
exit(EXIT_SUCCESS);
}
我看了这个,它应该在大多数情况下工作。然而,假设读入的字符串有连续7个字符的连续条纹。然后下一个字符串只有4条。前一个字符串的最后3个字符仍然会出现在'buf'我的想法中。如果你明白我的意思。 – user3268401
哦,是的,谢谢你指出。你可以添加另一行代码来清空数组,像这样buf [0] ='\ 0';在i = 0之后; –