fscanf() 只读入没有标点符号的字符

Question

我想从文本文件（在命令行中指定为参数的名称）中读入一些单词（在本例中为前 20 个单词）。当下面的代码运行时，我发现它也需要带字符的标点符号。

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

int main(int argc, char * argv[]){
int wordCap = 20;
int wordc = 0;
char** ptr = (char **) calloc (wordCap, sizeof(char*));
FILE *myFile = fopen (argv[1], "r");
if (!myFile) return 1;
rewind(myFile);
for (wordc = 0; wordc < wordCap; wordc++){
  ptr[wordc] = (char *)malloc(30 * sizeof( char ) );
  fscanf(myFile, "%s", ptr[wordc]);
  int length = strlen(ptr[wordc]);
  ptr[wordc][length] = '[=10=]';
   printf("word[%d] is %s\n", wordc,  ptr[wordc]);
}
 return 0;
}

当我遍历句子时："Once when a Lion was asleep a little Mouse began running up and down upon him;"，"him" 后面会跟一个分号。

我把fscanf()改成fscanf(myFile, "[a-z | A-Z]", ptr[wordc]);，把整个句子当成一个词。

我怎样才能改变它以获得正确的输出？

Answer 1

您可以接受分号，然后再将其删除，如下所示：

将单词存储到 ptr[wordc] 后:

i = 0;
while (i < strlen(ptr[wordc]))
{
    if (strchr(".;,!?", ptr[wordc][i])) //add any char you wanna delete to that string
        memmove(&ptr[wordc][i], &ptr[wordc][i + 1], strlen(ptr[wordc]) - i);
    else
        i++;
}
if (strlen(ptr[wordc]) > 0) // to not print any word that was just punctuations beforehand
    printf("word[%d] is %s\n", wordc,  ptr[wordc]);

我没有测试过这段代码，所以可能有错别字之类的。

或者您可以切换

fscanf(myFile, "%s", ptr[wordc]);

为了

fscanf(myFile, "%29[a-zA-Z]%*[^a-zA-Z]", ptr[wordc]);

只捕获字母。 29 限制了字的大小，所以你不会溢出，因为你只分配了 30 个字符的大小

fscanf() 只读入没有标点符号的字符

fscanf() to read in only characters with no punctuation marks

c

io

scanf