我想让它做的是，如果 word[count] == ',' 或 '.'去功能。当我在打印为 a 或 a 之前打印出 word[count] 的正确位置时。但它永远不会发挥作用。我什至尝试将其存储为 char 并进行比较，但是，当我将其存储为 char c 并打印时，我得到的只是一个 ' '.....(空白)这里的问题是什么。
#include<stdio.h>
void print_word(char word[], int count);
void special_char();
void fill(char word[], int count);
int main(){
FILE *fp;
char ch;
char word[50];
int count = 0;
int boolean_comma = 0;
fp = fopen("./../instr/lab1.dat","r");
if( fp == NULL) {
printf("ERROR");
}else
while(!feof(fp)){/*while pointer hasnt reached end of file continue loop*/
count++;
/*gets rid of the array jump, allowing it to reinitilize back to zero only at the beginning of each word*/
ch = fgetc(fp);
word[count] = ch; /*inset ch into array in coordinance with the count*/
/* printf("count: %i, char: %c\n",count,ch);*/
if(ch == ' ' || ch == '\n'){ /*Singals end of word*/
/*if last character is space and word[count-1] != ',' or a '.' */
/*then print out word normally, but if the [!=] above is correct */
/*And the count is equal to for fill the first 4 of the array */
/*with '*'s'|| but if word[count-1] == ',' or a '.' and the count*/
/*is == 5; then send to special char, which will fill first four */
/*of the array with '*'s' if the count is other than 5 sned it */
/*to the normal print_word function and allow it to print. */
if(ch == '\n'){count-=1;}
print_word(word,count); /*Sends array of words with count to print char by char*/
printf(" "); /*print space to separate words*/
count = 0;
}
}/* END while FEOF*/
fclose(fp);
return 0;
}//end main
/*This function will print all words, if the count is 4 it will fill with ASTRIK*/
/*The problem here is if the count is 4 and the fourth character is a special ch*/
/*to solve this problem check word[count] if it is equal to a ',' or '.' it */
/*must then bypass the normal for loop and go to a different one that will fill */
/*with the ASTRIKS and a COMMA or PERIOD */
void fill(char word[], int count)
{
int a;
for(a = 1; a <= count; a++){
word[a] = '*';
}/*END FOR*/
}/*END FILL*/
void print_word(char word[],int count)
{
int a;
char c = word[count];
count-=1;/*Gets rid of the count value which is a ' '*/
if(word[count] == ' '){count-=1;}/*gets rid of space*/
printf("count: %i ",count);
printf("[c:%c][w[:%c]\n",c,word[count]);
if(word[count] !=( ',' || '.')){
/*count-=1;*/
if( count == 4){
fill(word,count); /*Calls method fill*/
}/*END IF*/
}/*END IF*/
if(word[count] == (',' || '.')){
printf("here----------------\n");
if(count == 5){
fill(word,count); /*Calls the fill method with a -1*/
}/*END IF*/
}/*END ELSE IF*/
/*FOR LOOP wil end the function by printing edited word or unedited word*/
for(a = 1; a <= count; a++){
printf("%c",word[a]);
}/*END IF*/
printf("--%c", word[count]);
}/*end print_word*/
void speical_char(char word[],int count){
}/*end speical_char*/
最佳答案
在调用 feof() 之前，始终需要检查读取的返回值(fread()、fscanf() 或 fgetc())。
就像它进入循环的次数比你预期的多。如果出现读取错误，循环永远不会终止。
尝试:
int c;
while ((c = fgetc(fp)) != EOF) {
// do something with c
}
if (ferror(fp)) {
// handle the error, usually exit or return
} else {
// continue execution
}
关于c - 试图比较数组中的字符，但它永远不会捕获 C，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/32533461/