问题
I've some doubts about fgets. From what I know, it adds "\n" at the end of the string, and not "\0". So if I write this code:
fgets(buff,2,stdin);
printf("%s",buff);
So fgets reads two characters, I give as input "y", so buff should be "y\n". I'd expect printf to print y and add a line, while it prints "y" without adding a line. Can you explain why?
回答1:
char * fgets ( char * str, int num, FILE * stream );
Reads characters from input stream and stores them as a C string into str until (num-1) characters have been read or either a newline or the end-of-file is reached, whichever happens first.
A newline character makes fgets stop reading, but it is considered a valid character by the function and included in the string copied to str.
A terminating null character ('\0') is automatically appended after the characters copied to str.
回答2:
It must terminate the string, so yes it will always add '\0' at the end. However, fgets might not always add the newline, if it doesn't fit. I recommend this reference page for fgets.
回答3:
According to man page it clearly given :
char *fgets(char *s, int size, FILE *stream);
fgets() reads in at most one less than size characters from stream and stores them into the buffer pointed to by s. Reading stops after an EOF or a newline. If a newline is read, it is stored into the buffer. A '\0' is stored after the last character in the buffer.
来源：https://stackoverflow.com/questions/26339095/fgets-adds-0-or-n-at-the-end-of-the-input