我试图创建一个拼写检查程序,它接受一个输入文件,并通过搜索字典文件来确保每个单词是正确的。我面临的问题是,当我试图从输入文件中用空格分隔每个单词并将其放入char []中时,由于某种原因,使用"的单词会打印出来。
H0
i1
c0
h1
r2
i3
s4
!5
â0
1
2
h3
o4
w5
w6
â7
8
9
a0
r1
42
e3
y0
o1
u2
.3整数是我的索引
#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>
#include "dict.h"
int main(int argc, char *argv[] ) {
FILE *fdict,*input;
int i;
char ch;
/* the biggest posible word is 30 plus a possible of two " or ' characters and the null character. so the limit of the array is 33*/
char norm[33];
if ( argc < 3 ) /* argc should be 3 for correct execution*/
{
fprintf(stderr,"1 or 2 Files were missing.");
exit(1);
}
if ( argc > 3 ){
fprintf(stderr,"too many Arguments");
exit(1);
}
/* We assume argv[1] and agrv[2] are filenames to open*/
fdict = fopen( argv[1], "r" );/* file pointer for the dictionary file*/
input = fopen( argv[2], "r" );/*file pointer for the input file*/
/* fopen returns NULL on failure */
if ( fdict == NULL ){
fprintf(stderr,"Could not open file: %s\n", argv[1] );/*checks to make sure the dictionary file can be opened*/
exit(1);
}
if ( input == NULL ){
fprintf(stderr,"Could not open file: %s\n", argv[2] );/*checks to make sure the input file can be opened*/
exit(1);
}
/* Read one character at a time from file, stopping at EOF, which
indicates the end of the file. Note that the idiom of "assign
to a variable, check the value" used below works because
the assignment statement evaluates to the value assigned. */
while ( ( ch = fgetc( input ) ) != EOF ) {
char word[33] = "";/* resets the array*/
for ( i = 0; !isspace( ch ) ; i++ ){
word[i] = ch;
printf("%c%d\n",ch,i);/* checking to see what is wrong with the index*/
ch = fgetc( input );
}
}
fclose( fdict );
fclose( input );
return 0;
}我的输入看起来是:
Hi chris! “howw” are you.发布于 2014-09-24 16:00:05
"与“或”不一样。(3个不同的引号)基于不同的编码,这三个字符使用不同的char序列来表示它们,而代码一次只打印一个char。
建议只使用simpe引号标记"。
一个简单的或程序员的文本编辑器就可以了。避免使用可能会引入非ASCII引号的字处理器,直到代码准备就绪(@n.m.)
https://stackoverflow.com/questions/26021322
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