fscanf（）仅读取没有标点符号的字符

Question

I would like to read in some words (in this example first 20) from a text file (name specified as an argument in the command line). As the below code runs, I found it takes punctuation marks with characters too.

#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] = '\0';
   printf("word[%d] is %s\n", wordc,  ptr[wordc]);
}
 return 0;
}

As I pass through the sentence: "Once when a Lion was asleep a little Mouse began running up and down upon him;", "him" will be followed with a semicolon.

I changed the fscanf() to be fscanf(myFile, "[az | AZ]", ptr[wordc]); , it takes the whole sentence as a word.

How can I change it to make the correct output?

Answer 1

You could accept the semi-colon and then remove it latter, like so:

after you've stored the word in ptr[wordc]:

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]);

I haven't tested this code, so there might be a typo or something in it.

Alternatively you could switch

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

for

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

to capture only letters. the 29 limits word size so you don't get overflow since you're allocating size for only 30 chars