fgetc

when the following code is compiled it goes into an infinite loop: int main() { unsigned char ch; FILE *fp; fp = fopen("abc","r"); if(fp==NULL) { printf("Unable to Open"); exit(1); } while((ch = fgetc(fp))!=EOF) printf("%c",ch); fclose(fp); printf("\n",ch); return 0; } The gcc Compiler also gives warning on compilation abc.c:13:warning: comparison is always true due to limited range of data type the code runs fine when unsigned char is replaced by char or int as expected i.e. it terminates. But the code also runs fine for unsigned int as well. as i have i have read in EOF is defines as -1 in

In the book Linux System Programming I have read some like this: fgetc returns the character read as an unsigned char cast to an int or EOF on end of file or error. A common error using fgetc is: char c; if ((c = fgetc()) != EOF) {...} The right version of this code is: int c; if ((c = fgetc()) != EOF) { printf("%c", (char)c); ... } So, why can't I cast a return value to char before comparing with EOF ? Why do I have to compare EOF exactly with int ? As EOF defined as -1 , isn't it normally casted to char ? Are there platforms/compilers where it is not true? Cacho Santa You can't cast the