也就是说用C文件计数如Linux wc命令 [英] words counting in file like linux wc command in C
问题描述
我试图写一些东西,就像Linux命令厕所来算的话,新的生产线和字节中的任何类型的文件,我可以只使用读取C函数。我写了这个code和我得到了新行和字节的正确值,但我不计的话得到正确的值。
字节INT = 0;
INT字= 0;
INT NEWLINE = 0;
炭缓冲液[1];
INT文件=打开(MYFILE,O_RDONLY);
如果(文件== -1){
的printf(找不到:%S \\ n,MYFILE);
}
其他{
最后烧焦='C';
而(读(文件,缓冲液,1)== 1){
字节++;
如果(缓冲[0] ==''和;&安培;最后=!''和;&安培;最后='\\ n'!){
也就是说++;
}
否则,如果(缓冲[0] =='\\ n'){
NEWLINE ++;
如果(最后='!'和;&安培;最后!='\\ n'){
也就是说++;
}
}
最后=缓冲器[0];
}
的printf(%d个%D%S \\ n,换行,字数,字节数,MYFILE);
}
您应该反转的逻辑。而不是寻找一个空间,并增加你的字数,寻找一个非空间来增加字数。此外,它可以帮助使用状态变量与看最后一个字符:
INT主要(无效)
{
为const char * MYFILE =的test.txt;
INT字节= 0;
INT字= 0;
INT NEWLINE = 0;
炭缓冲液[1];
INT文件=打开(MYFILE,O_RDONLY);
枚举状态{WHITESPACE,WORD};
INT状态= WHITESPACE;
如果(文件== -1){
的printf(找不到:%S \\ n,MYFILE);
}
其他{
焦炭最后='';
而(读(文件,缓冲液,1)== 1)
{
字节++;
如果(缓冲[0] ==''||缓冲区[0] =='\\ t')
{
状态= WHITESPACE;
}
否则,如果(缓冲[0] =='\\ n')
{
NEWLINE ++;
状态= WHITESPACE;
}
其他
{
如果(状态== WHITESPACE)
{
也就是说++;
}
状态= WORD;
}
最后=缓冲器[0];
}
的printf(%d个%D%S \\ n,换行,字数,字节数,MYFILE);
}}
看来,厕所有一定的逻辑相对于标点符号不是的话,那这个code不处理。
I am trying to write something that works like the Linux command wc to count words, new lines and bytes in any kind of files and i can only use the C function read. I have written this code and i am getting the correct values for newlines and bytes but i am not getting the correct value for counted words.
int bytes = 0;
int words = 0;
int newLine = 0;
char buffer[1];
int file = open(myfile,O_RDONLY);
if(file == -1){
printf("can not find :%s\n",myfile);
}
else{
char last = 'c';
while(read(file,buffer,1)==1){
bytes++;
if(buffer[0]==' ' && last!=' ' && last!='\n'){
words++;
}
else if(buffer[0]=='\n'){
newLine++;
if(last!=' ' && last!='\n'){
words++;
}
}
last = buffer[0];
}
printf("%d %d %d %s\n",newLine,words,bytes,myfile);
}
You should reverse your logic. Rather than look for a space, and increment your word count, look for a non-space to increment the word count. Also, it can help to use a state variable versus looking at the last char:
int main(void)
{
const char *myfile = "test.txt";
int bytes = 0;
int words = 0;
int newLine = 0;
char buffer[1];
int file = open(myfile,O_RDONLY);
enum states { WHITESPACE, WORD };
int state = WHITESPACE;
if(file == -1){
printf("can not find :%s\n",myfile);
}
else{
char last = ' ';
while (read(file,buffer,1) ==1 )
{
bytes++;
if ( buffer[0]== ' ' || buffer[0] == '\t' )
{
state = WHITESPACE;
}
else if (buffer[0]=='\n')
{
newLine++;
state = WHITESPACE;
}
else
{
if ( state == WHITESPACE )
{
words++;
}
state = WORD;
}
last = buffer[0];
}
printf("%d %d %d %s\n",newLine,words,bytes,myfile);
}
}
It appears that wc has some logic with respect to punctuation characters not being words, that this code does not handle.
这篇关于也就是说用C文件计数如Linux wc命令的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!