fgetc().....Don't Quite Get This:

fgetc() returns an int, right? Then why is it I can assign c, a character, to the result of fgetc(fp)? ...It doesn't make any sense to me.

Code:

#include <stdio.h>

int main(void){
    FILE *fp = fopen("myFile.txt","r");
    if(!fp){
            printf("Error opening file.");
    }
    char c;
    int n=0;
    while(c!=EOF){
                 c = fgetc(fp);
                  printf("%c",c);
                  if(c == '$'){
                       n++;
                  }
    }
    printf("Your file has %d$s.\n",n);
    getchar();
    return 0;
}

Well... int is converted to char implicitly. But char is not guaranteed to be capable of holding EOF. So using char to hold the return value of fgetc is wrong!

Use an int to get the return value of fgetc, check if it is EOF. Once you've established that it's not EOF, you can safely put it in a char.

Char values can be converted to ascii values if you like. The compiler does the job for you. But when you call fgetc() the compiler expects char or an int representation of that char. For example:

int c;
c=fgetc(); -->and you press letter 'A' on your keyboar , the value of c becomes 65.See the ascii table.

or if you cast any char to int you get it's number in the ascii table.

Code:

  char c='A';
   printf("%d",(int)c);

In the above example I wouldn't have to cast char to int sice we already have formated output using %d...

You don't need to cast to int:

Code:

char c='A';
printf("%d", c);

The above is fine, since c will be promoted to int when passed to printf.

Originally Posted by cwr

You don't need to cast to int:

Code:

char c='A';
printf("%d", c);

The above is fine, since c will be promoted to int when passed to printf.

Interpreted...

Originally Posted by xeddiex

Interpreted...

What exactly do you mean?

I think that means "point taken and understood."