A beginners' guide away from scanf() Felix Palmen felix@palmen-it.de

This document is for you if you started to learn programmming in C . Chances are you follow a course and the method to read some input you were taught is to use the scanf() function.

0. What's wrong with scanf() ?

Nothing. And, chances are, everything for your usecase. This document attempts to make you understand why. So here's the very first rule about scanf() :

Rule 0: Don't use scanf() . (Unless, you know exactly what you do.)

But before presenting some alternatives for common usecases, let's elaborate a bit on the knowing what you do part.

1. I want to read a number from the user

Here is a classic example of scanf() use (and, misuse) in a beginner's program:

1 2 3 4 5 6 7 8 9 #include <stdio.h> int main( void ) { int a; printf( "enter a number: " ); scanf( "%d" , &a); printf( "You entered %d.

" , a); }

As you probably know, %d is the conversion for an integer, so this program works as expected:

$ ./example1 enter a number: 42 You entered 42.

Or does it?

$ ./example1 enter a number: abcdefgh You entered 38.

Oops. Where does the value 38 come from?

The answer is: This could be any value, or the program could just crash. A crashing program in just two lines of code is quite easy to create in C . scanf() is asked to convert a number, and the input doesn't contain any numbers, so scanf() converts nothing. As a consequence, the variable a is never written to and using the value of an uninitialized variable in C is undefined behavior.

Undefined behavior in C C is a very low-level language and one consequence of that is the following: Nothing will ever stop you from doing something completely wrong. Many languages, especially those for some managed environment like Java or C# actually stop you when you do things that are not allowed, say, access an array element that does not exist. C doesn't. As long as your program is syntactically correct, the compiler won't complain. If you do something forbidden in your program, C just calls the behavior of your program undefined. This formally allows anything to happen when running the program. Often, the result will be a crash or just output of "garbage" values, as seen above. But if you're really unlucky, your program will seem to work just fine until it gets some slightly different input, and by that time, you will have a really hard time to spot where exactly your program is undefined. Therefore avoid undefined behavior by all means!. On a side note, undefined behavior can also cause security holes. This has happened a lot in practice.

Now that we know the program is broken, let's fix it. Because scanf() returns how many items were converted successfully, the next obvious idea is just to retry the "number input" in case the user entered something else:

1 2 3 4 5 6 7 8 9 10 11 12 13 #include <stdio.h> int main( void ) { int a; printf( "enter a number: " ); while (scanf( "%d" , &a) != 1 ) { // input was not a number, ask again: printf( "enter a number: " ); } printf( "You entered %d.

" , a); }

Let's test:

$ ./example2 enter a number: abc enter a number: enter a number: enter a number: enter a number: enter a number: enter a number: enter a number: enter a number: enter a number: enter a number: enter a number: enter a number: enter a number: enter a number: enter a number: enter a number: enter a number: enter a number: enter a number: enter a number: enter a number: enter a number: enter a number: enter a number: enter a number: enter a number: enter a number: enter a number: enter a number: enter a number: enter a number: enter a number: enter a number: enter a number: enter a number: enter a number: enter a number: enter a number: enter a number: enter a number: enter a number: enter a number: enter a number: enter a number: enter a number: enter a number: enter a number: enter a number: enter a number: enter a number: enter a number: enter a number: ^C

stooooop! Ok, we managed to interrupt this madness with Ctrl+C but why did that happen?

Here's a rule:

Rule 1: scanf() is not for reading input, it's for parsing input.

The first argument to scanf() is a format string, describing what scanf() should parse. The important thing is: scanf() never reads anything it cannot parse. In our example, we tell scanf() to parse a number, using the %d conversion. Sure enough, abc is not a number, and as a consequence, abc is not even read. The next call to scanf() will again find our unread input and, again, can't parse it.

Chances are you find some examples saying "let's just flush the input before the next call to scanf() ":

fflush(stdin); // <- never do that!

Forget about this idea immediately, please.

You'd expect this to clear all unread input, and indeed, some systems will do just that. But according to C , flushing an input stream is undefined behavior, and this should now ring a bell. And yes, there are a lot of systems that won't clear the input when you attempt to flush stdin .

So, the only way to clear unread input is by reading it. Of course, we can make scanf() read it, using a format string that parses any string. Sounds easy.

2. I want to read a string from the user

Let's consider another classic example of a beginner's program, trying to read a string from the user:

1 2 3 4 5 6 7 8 9 #include <stdio.h> int main( void ) { char name[ 12 ]; printf( "What's your name? " ); scanf( "%s" , name); printf( "Hello %s!

" , name); }

As %s is for strings, this should work with any input:

$ ./example3 What's your name? Paul Hello Paul! $ ./example3 What's your name? Christopher-Joseph-Montgomery Segmentation fault $

Well, now we have a buffer overflow. You might get Segmentation fault on a Linux system, any other kind of crash, maybe even a "correctly" working program, because, once again, the program has undefined behavior.

The problem here is: %s matches any string, of any length, and scanf() has no idea when to stop reading. It reads as long as it can parse the input according to the format string, so it writes a lot more data to our name variable than the 12 characters we declared for it.

Buffer overflows in C A buffer overflow is a specific kind of undefined behavior resulting from a program that tries to write more data to an (array) variable than this variable can hold. Although this is undefined, in practice it will result in overwriting some other data (that happens to be placed after the overflowed buffer in memory) and this can easily crash the program. One particularly dangerous result of a buffer overflow is overwriting the return address of a function. The return address is used when a function exits, to jump back to the calling function. Being able to overwrite this address ultimately means that a person with enough knowledge about the system can cause the running program to execute any other code supplied as input. This problem has led to many security vulnerabilities; imagine you can make for example a webserver written in C execute your own code by submitting a specially tailored request...

So, here's the next rule:

Rule 2: scanf() can be dangerous when used carelessly. Always use field widths with conversions that parse to a string (like %s ).

The field width is a number preceeding the conversion specifier. It causes scanf() to consider a maximum number of characters from the input when parsing for this conversion. Let's demonstrate it in a fixed program:

1 2 3 4 5 6 7 8 9 #include <stdio.h> int main( void ) { char name[ 40 ]; printf( "What's your name? " ); scanf( "%39s" , name); printf( "Hello %s!

" , name); }

We also increased the buffer size, because there might be really long names.

There's an important thing to notice: Although our name has room for 40 characters, we instruct scanf() not to read more than 39. This is because a string in C always needs a 0 byte appended to mark the end. When scanf() is finished parsing into a string, it appends this byte automatically, and there must be space left for it.

So, this program is now safe from buffer overflows. Let's try something different:

$ ./example4 What's your name? Martin Brown Hello Martin!

Well, that's ... outspoken. What happens here? Reading some scanf() manual, we would find that %s parses a word, not a string, for example I found the following wording:

s: Matches a sequence of non-white-space characters

A white-space in C is one of space, tab ( \t ) or newline (

).

Rule 3: Although scanf() format strings can look quite similar to printf() format strings, they often have slightly different semantics. (Make sure to read the fine manual)

The general problem with parsing "a string" from an input stream is: Where does this string end? With %s , the answer is at the next white-space. If you want something different, you can use %[ :

%[a-z] : parse as long as the input characters are in the range a - z .

: parse as long as the input characters are in the range - . %[ny] : parse as long as the input characters are y or n .

: parse as long as the input characters are or . %[^.] : The ^ negates the list, so this means parse as long as there is no . in the input.

We could change the program, so anything until a newline will be parsed into our string:

1 2 3 4 5 6 7 8 9 #include <stdio.h> int main( void ) { char name[ 40 ]; printf( "What's your name? " ); scanf( "%39[^

]" , name); printf( "Hello %s!

" , name); }

It might get a bit frustrating, but this is again a program with possible undefined behavior, see what happens when we just press Enter :

$ ./example5 What's your name? Hello ÿ¦e!

Here's another sentence from a scanf() manual, from the section describing the [ conversion:

The usual skip of leading white space is suppressed.

With many conversions, scanf() automatically skips whitespace characters in the input, but with some, it doesn't. Here, our newline from just pressing enter isn't skipped, and it doesn't match for our conversion that explicitly excludes newlines. The result is: scanf() doesn't parse anything, our name remains uninitialized.

One way around this is to tell scanf() to skip whitespace: If the format string contains any whitespace, it matches any number of whitespace characters in the input, including no whitespace at all. Let's use this to skip whitespace the user might enter before entering his name:

1 2 3 4 5 6 7 8 9 10 #include <stdio.h> int main( void ) { char name[ 40 ]; printf( "What's your name? " ); scanf( " %39[^

]" , name); // ^ note the space here, matching any whitespace printf( "Hello %s!

" , name); }

Yes, this program works and doesn't have any undefined behavior*), but I guess you don't like very much that nothing at all happens when you just press enter, because scanf() is skipping it and continues to wait for input that can be matched.

*) actually, this isn't even entirely true. This program still has undefined behavior for empty input. You could force this on a Linux console hitting Ctrl+D for example. So, it's again an example for code you should not write.

3. Ok, I just want to read some input from the user

There are several functions in C for reading input. Let's have a look at one that's probably most useful to you: fgets() .

fgets() does a simple thing, it reads up to a given maximum number of characters, but stops at a newline, which is read as well. In other words: It reads a line of input.

This is the function signature:

char *fgets( char *str, int n, FILE *stream)

There are two very nice things about this function for what we want to do:

The parameter for the maximum length accounts for the necessary 0 byte, so we can just pass the size of our variable.

byte, so we can just pass the size of our variable. The return value is either a pointer to str or NULL if, for any reason, nothing was read.

So let's rewrite this program again:

1 2 3 4 5 6 7 8 9 10 11 #include <stdio.h> int main( void ) { char name[ 40 ]; printf( "What's your name? " ); if (fgets(name, 40 , stdin)) { printf( "Hello %s!

" , name); } }

I assure you this is safe, but it has a little flaw:

$ ./example7 What's your name? Bob Hello Bob !

Of course, this is because fgets() also reads the newline character itself. But the fix is simple as well: We use strcspn() to get the index of the newline character if there is one and overwrite it with 0 . strcspn() is declared in string.h , so we need a new #include :

1 2 3 4 5 6 7 8 9 10 11 12 13 #include <stdio.h> #include <string.h> int main( void ) { char name[ 40 ]; printf( "What's your name? " ); if (fgets(name, 40 , stdin)) { name[strcspn(name, "

" )] = 0 ; printf( "Hello %s!

" , name); } }

Let's test it:

$ ./example8 What's your name? Bob Belcher Hello Bob Belcher!

4. How would I get numbers without scanf() ?

There are many functions for converting a string to a number in C . A function that's used quite often is atoi() , the name means anything to integer. It returns 0 if it can't convert the string. Let's try to rewrite the broken example 2 using fgets() and atoi() . atoi() is declared in stdlib.h .

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 #include <stdio.h> #include <stdlib.h> int main( void ) { int a; char buf[ 1024 ]; // use 1KiB just to be sure do { printf( "enter a number: " ); if (!fgets(buf, 1024 , stdin)) { // reading input failed, give up: return 1 ; } // have some input, convert it to integer: a = atoi(buf); } while (a == 0 ); // repeat until we got a valid number printf( "You entered %d.

" , a); }

So, trying this out:

$ ./example9 enter a number: foo enter a number: bar enter a number: 15x You entered 15.

Well, not bad so far. But what if we want to allow an actual 0 to be entered? We can't tell whether atoi() returns 0 because it cannot convert anything or because there was an actual 0 in the string. Also, ignoring the extra x when we input 15x may not be what we want.

atoi() is good enough in many cases, but if you want better error checking, there's an alternative: strtol() :

long int strtol( const char *nptr, char **endptr, int base);

This looks complicated. But it isn't:

endptr is set to point at the first character that couldn't be converted. So you have a way to check whether the whole string was converted.

base allows you to specify any base you expect a number in. Most of the time, this will be 10 , but you could give 16 here for parsing hexadecimal or 2 for parsing binary.

, but you could give here for parsing hexadecimal or for parsing binary. strtol() even sets errno , so you can check whether a number was too small or too big for conversion.

Now let's use this instead of atoi() (note it returns a long int ), making use of every possibility to detect errors:

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 #include <stdio.h> #include <stdlib.h> #include <errno.h> int main( void ) { long a; char buf[ 1024 ]; // use 1KiB just to be sure int success; // flag for successful conversion do { printf( "enter a number: " ); if (!fgets(buf, 1024 , stdin)) { // reading input failed: return 1 ; } // have some input, convert it to integer: char *endptr; errno = 0 ; // reset error number a = strtol(buf, &endptr, 10 ); if (errno == ERANGE) { printf( "Sorry, this number is too small or too large.

" ); success = 0 ; } else if (endptr == buf) { // no character was read success = 0 ; } else if (*endptr && *endptr != '

' ) { // *endptr is neither end of string nor newline, // so we didn't convert the *whole* input success = 0 ; } else { success = 1 ; } } while (!success); // repeat until we got a valid number printf( "You entered %ld.

" , a); }

And again, let's try:

$ ./example10 enter a number: 565672475687456576574 Sorry, this number is too small or too large. enter a number: ggggg enter a number: 15x enter a number: 0 You entered 0.

This looks really good, doesn't it? If you want to know more, I suggest you read on similar functions like atof() , strtoll() , strtod() etc.

5. But can I fix the examples with scanf() as well?

Yes, you can. Here's a last rule:

Rule 4: scanf() is a very powerful function. (and with great power comes great responsibility ...)

A lot of parsing work can be done with scanf() in a very concise way, which can be very nice, but it also has many pitfalls and there are tasks (such as reading a line of input) that are much simpler to accomplish with a simpler function. Make sure you understand the rules presented here, and if in doubt, read the scanf() manual precisely.

That being said, here's an example how to read a number with retries using scanf() :

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 #include <stdio.h> int main( void ) { int a; int rc; printf( "enter a number: " ); while ((rc = scanf( "%d" , &a)) == 0 ) // Neither success (1) nor EOF { // clear what is left, the * means only match and discard: scanf( "%*[^

]" ); // input was not a number, ask again: printf( "enter a number: " ); } if (rc == EOF) { printf( "Nothing more to read - and no number found

" ); } else { printf( "You entered %d.

" , a); } }

It's not as nice as the version using strtol() above, because there is no way to tell scanf() not to skip whitespace for %d -- so if you just hit Enter , it will still wait for your input -- but it works and it's a really short program.

For the sake of completeness, if you really really want to get a line of input using scanf() , of course this can be done safely as well:

1 2 3 4 5 6 7 8 9 10 11 #include <stdio.h> int main( void ) { char name[ 40 ]; printf( "What's your name? " ); if (scanf( "%39[^

]%*c" , name) == 1 ) // We expect exactly 1 conversion { printf( "Hello %s!

" , name); } }