Problem using Scanf? - c

Why does scanf give a max value in case of "int" but crash the program in case of "char" when the limit is exceeded?
#include<stdio.h>
main(){
int a;
char ch[10];
scanf("%d",&a);
printf("%d",a);
scanf("%s",ch);
printf("%s",ch);
}

It crashes your program in this case because scanf() has an inherently unsafe interface for strings. It has no way of knowing that your parameter ch is an array large enough to hold a 9-character string plus its terminating nul. As a result, it is perfectly happy to keep reading characters from stdin and storing them in memory well past the end of the array.
If your program crashes, you are actually lucky. In the worst case, the attacker has used a carefully crafted string to manipulate the content of the stack in such a way that he has gained control of your computer. This is an example of a buffer overflow attack. It sounds unlikely, but it has been documented to occur on a large number of occasions.
Used for only numbers, scanf is generally safe enough, but it is not very good at handling errors in the input. As a result, it is usually a good idea to use something like fgets() to read the input (it has a buffer length parameter to control overflow) and sscanf() to parse from that buffer, testing its return values for sanity as you go.
Edit: As the comment from R points out, I overstated the dangers inherent to the scanf interface. With care to correctly use the field width specifier on all strings, scanf becomes safer. But then you take responsibility for guaranteeing that the specified width does fit within the buffer. For the example, you should write scanf("%9s",ch); because your buffer was declared to be ten bytes long and you need room for the terminating nul.
Note that you should also be testing the return value from scanf(). It returns the number of fields it successfully matched, or EOF if an I/O error occurred. It might return 0 if the user entered "abc" when it expected a number, for instance.

Because you're not reading a character, you're reading a string. And scanf does not "know" that you only have space for 10 characters (including the null). This is the joy of C programming.
You can protect yourself in this case by adding a width modifier:
%9s

Related

Why does scanf accept more characters than there is room for in the buffer?

See the following code:
int main()
{
char test[3];
scanf("%s", test);
__fpurge(stdin);
printf("%s", test);
}
The program should record only 3 characters, but when I type, for example, 8 characters, the program records all 8! This should not happen. The correct would record 3 characters, because the scanf do it?
scanf accepts more data than you can fit in test because you allow it to do so by using %s without a limit. This is dangerous, and must be avoided in production code.
Replace %s with %3s to fix this problem. If you want to read three characters, test must be four-characters wide to accommodate null terminator:
char test[4];
scanf("%3s", test);
When you pass test to scanf(), you are passing nothing but a pointer to the first character of your buffer, so scanf() has no idea how large your buffer is. It will happily accept as many characters as you type, and it will store them all in there. So, when you type more than 2 characters, you are causing scanf() to write characters (plus the zero asciiz terminator character) past the end of your buffer. Normally, what is to be expected in such a case is a program crash.
The fact that you did not experience a crash is largely coincidence, what is probably happening is that the compiler has allocated room for more than 3 characters in the stack due to alignment considerations, possibly room for 8 characters or more. If you type enough characters, your program will surely crash.
For this reason, this usage of scanf() is considered completely unsafe. One should never use scanf() like that when doing any serious coding. Instead you should specify the width of your string, like this: "%2s". (Note that you must specify a number which is smaller than the size of your buffer by one, in order to account for the zero asciiz terminator character that will be automatically appended by scanf().)

inputting a character string using scanf()

I started learning about inputting character strings in C. In the following source code I get a character array of length 5.
#include<stdio.h>
int main(void)
{
char s1[5];
printf("enter text:\n");
scanf("%s",s1);
printf("\n%s\n",s1);
return 0;
}
when the input is:
1234567891234567, and I've checked it's working fine up to 16 elements(which I don't understand because it is more than 5 elements).
12345678912345678, it's giving me an error segmentation fault: 11 (I gave 17 elements in this case)
123456789123456789, the error is Illegal instruction: 4 (I gave 18 elements in this case)
I don't understand why there are different errors. Is this the behavior of scanf() or character arrays in C?. The book that I am reading didn't have a clear explanation about these things. FYI I don't know anything about pointers. Any further explanation about this would be really helpful.
Is this the behavior of scanf() or character arrays in C?
TL;DR - No, you're facing the side-effects of undefined behavior.
To elaborate, in your case, against a code like
scanf("%s",s1);
where you have defined
char s1[5];
inputting anything more than 4 char will cause your program to venture into invalid memory area (past the allocated memory) which in turn invokes undefined behavior.
Once you hit UB, the behavior of the program cannot be predicted or justified in any way. It can do absolutely anything possible (or even impossible).
There is nothing inherent in the scanf() which stops you from reading overly long input and overrun the buffer, you should keep control on the input string scanning by using the field width, like
scanf("%4s",s1); //1 saved for terminating null
The scanf function when reading strings read up to the next white-space (e.g. newline, space, tab etc.), or the "end of file". It has no idea about the size of the buffer you provide it.
If the string you read is longer than the buffer provided, then it will write out of bounds, and you will have undefined behavior.
The simplest way to stop this is to provide a field length to the scanf format, as in
char s1[5];
scanf("%4s",s1);
Note that I use 4 as field length, as there needs to be space for the string terminator as well.
You can also use the "secure" scanf_s for which you need to provide the buffer size as an argument:
char s1[5];
scanf_s("%s", s1, sizeof(s1));

Does the C standard guarantee buffers are not touched past their null terminator?

In the various cases that a buffer is provided to the standard library's many string functions, is it guaranteed that the buffer will not be modified beyond the null terminator? For example:
char buffer[17] = "abcdefghijklmnop";
sscanf("123", "%16s", buffer);
Is buffer now required to equal "123\0efghijklmnop"?
Another example:
char buffer[10];
fgets(buffer, 10, fp);
If the read line is only 3 characters long, can one be certain that the 6th character is the same as before fgets was called?
The C99 draft standard does not explicitly state what should happen in those cases, but by considering multiple variations, you can show that it must work a certain way so that it meets the specification in all cases.
The standard says:
%s - Matches a sequence of non-white-space characters.252)
If no l length modifier is present, the corresponding argument shall be a
pointer to the initial element of a character array large enough to accept the
sequence and a terminating null character, which will be added automatically.
Here's a pair of examples that show it must work the way you are proposing to meet the standard.
Example A:
char buffer[4] = "abcd";
char buffer2[10]; // Note the this could be placed at what would be buffer+4
sscanf("123 4", "%s %s", buffer, buffer2);
// Result is buffer = "123\0"
// buffer2 = "4\0"
Example B:
char buffer[17] = "abcdefghijklmnop";
char* buffer2 = &buffer[4];
sscanf("123 4", "%s %s", buffer, buffer2);
// Result is buffer = "123\04\0"
Note that the interface of sscanf doesn't provide enough information to really know that these were different. So, if Example B is to work properly, it must not mess with the bytes after the null character in Example A. This is because it must work in both cases according to this bit of spec.
So implicitly it must work as you stated due to the spec.
Similar arguments can be placed for other functions, but I think you can see the idea from this example.
NOTE:
Providing size limits in the format, such as "%16s", could change the behavior. By the specification, it would be functionally acceptable for sscanf to zero out a buffer to its limits before writing the data into the buffer. In practice, most implementations opt for performance, which means they leave the remainder alone.
When the intent of the specification is to do this sort of zeroing out, it is usually explicitly specified. strncpy is an example. If the length of the string is less than the maximum buffer length specified, it will fill the rest of the space with null characters. The fact that this same "string" function could return a non-terminated string as well makes this one of the most common functions for people to roll their own version.
As far as fgets, a similar situation could arise. The only gotcha is that the specification explicitly states that if nothing is read in, the buffer remains untouched. An acceptable functional implementation could sidestep this by checking to see if there is at least one byte to read before zeroing out the buffer.
Each individual byte in the buffer is an object. Unless some part of the function description of sscanf or fgets mentions modifying those bytes, or even implies their values may change e.g. by stating their values become unspecified, then the general rule applies: (emphasis mine)
6.2.4 Storage durations of objects
2 [...] An object exists, has a constant address, and retains its last-stored value throughout its lifetime. [...]
It's this same principle that guarantees that
#include <stdio.h>
int a = 1;
int main() {
printf ("%d\n", a);
printf ("%d\n", a);
}
attempts to print 1 twice. Even though a is global, printf can access global variables, and the description of printf doesn't mention not modifying a.
Neither the description of fgets nor that of sscanf mentions modifying buffers past the bytes that actually were supposed to be written (except in the case of a read error), so those bytes don't get modified.
The standard is somewhat ambiguous on this, but I think a reasonable reading of it is that the answer is: yes, it's not allowed to write more bytes to the buffer than it read+null. On the other hand, a stricter reading/interpretation of the text could conclude that the answer is no, there's no guarantee. Here's what a publicly avaialble draft says about fgets.
char *fgets(char * restrict s, int n, FILE * restrict stream);
The fgets function reads at most one less than the number of characters specified by n from the stream pointed to by stream into the array pointed to by s. No additional characters are read after a new-line character (which is retained) or after end-of-file. A null character is written immediately after the last character read into the array.
The fgets function returns s if successful. If end-of-file is encountered and no characters have been read into the array, the contents of the array remain unchanged and a null pointer is returned. If a read error occurs during the operation, the array contents are indeterminate and a null pointer is returned.
There's a guarantee about how much it is supposed to read from the input, i.e. stop reading at newline or EOF and not read more than n-1 bytes. Although nothing is said explicitly about how much it's allowed to write to the buffer, the common knowledge is that fgets's n parameter is used to prevent buffer overflows. It's a little strange that the standard uses the ambiguous term read, which may not necessarily imply that gets can't write to the buffer more than n bytes, if you want to nitpick on the terminology it uses. But note that the same "read" terminology is used about both issues: the n-limit and the EOF/newline limit. So if you interpret the n-related "read" as a buffer-write limit, then [for consistency] you can/should interpret the other "read" the same way, i.e. not write more than what it read when string is shorter than the buffer.
On the other hand, if you distinguish between the uses of the phrase-verb "read into" (="write") and just "read", then you can't read the committee's text the same way. You are guaranteed that it won't "read into" (="write to") the array more than n bytes, but if the input string is terminated sooner by newline or EOF you're only guaranteed the rest (of the input) won't be "read", but whether that implies in won't be "read into" (="written to") the buffer is unclear under this stricter reading. The crucial issue is keyword is "into", which is elided, so the problem is whether the completion given by me in brackets in the following modified quote is the intended interpretation:
No additional characters are read [into the array] after a new-line character (which is retained) or after end-of-file.
Frankly a single postcondition stated as a formula (and would be pretty short in this case) would have been a lot more helpful than the verbiage I quoted...
I can't be bothered to try and analyze their writeup about the *scanf family, because I suspect it's going to be even more complicated given all the other things that happen in those functions; their writeup for fscanf is about five pages long... But I suspect a similar logic applies.
is it guaranteed that the buffer will not be modified beyond the null
terminator?
No, there's no guarantee.
Is buffer now required to equal "123\0efghijklmnop"?
Yes. But that's only because you've used correct parameters to your string related functions. Should you mess up buffer length, input modifiers to sscanf and such, then you program will compile. But it will most likely fail during runtime.
If the read line is only 3 characters long, can one be certain that the 6th character is the same as before fgets was called?
Yes. Once fgets() figures you have a 3 character input string it stores the input in the provided buffer, and it doesn't care about the reset of provided space at all.
Is buffer now required to equal "123\0efghijklmnop"?
Here buffer is just consists of 123 string guaranteed terminating at NUL.
Yes the memory allocated for array buffer will not get de-allocated, however you are making sure/restricting your string buffer can atmost only have 16 char elements which you can read into it at any point of time. Now depends whether you write just a single char or maximum what buffer can take.
For example:
char buffer[4096] = "abc";`
actually does something below,
memcpy(buffer, "abc", sizeof("abc"));
memset(&buffer[sizeof("abc")], 0, sizeof(buffer)-sizeof("abc"));
The standard insists that if any part of char array is initialized that is all it consists of at any moment until obeying its memory boundary.
There are no any guarantees from standard, which is why the functions sscanf and fgets are recommended to be used (with respect to the size of the buffer) as you show in your question (and using of fgets is considered preferable compared with gets).
However, some standard functions use null-terminator in their work, e.g. strlen (but I suppose you ask about string modification)
EDIT:
In your example
fgets(buffer, 10, fp);
untouching characters after 10-th is guaranteed (content and length of buffer will not be considered by fgets)
EDIT2:
Moreover, when using fgets keep in mind that '\n' will be stored in the buffers. e.g.
"123\n\0fghijklmnop"
instead of expected
"123\0efghijklmnop"
Depends on the function in use (and to a lesser degree its implementation). sscanf will start writing when it encounters its first non-whitespace character, and continue writing until its first whitespace character, where it will add a finishing 0 and return. But a function like strncpy (famously) zeroes out the rest of the buffer.
There is however nothing in the C standard which mandates how these functions behave.

difference betweent printf and gets

I am beginner for programming.I referred books of C programming,but i am confused.
1.) What's the difference betweent printf and gets?
I believe gets is simpler and doesn't have any formats?
printf
The printf function writes a formatted string to the standard output. A formatted string is the result of replacing placeholders with their values. This sounds a little complicated but it will become very clear with an example:
printf("Hello, my name is %s and I am %d years old.", "Andreas", 22);
Here %s and %d are the placeholders, that are substituted with the first and second argument. You should read on the man page (linked above) the list of placeholders and their options, but the ones you'll run into most often are %d (a number) and %s (a string).
Making sure that the placeholder arguments match their type is extremely important. For example, the following code will result in undefined behavior (meaning that anything can happen: the program may crash, it may work, it may corrupt data, etc):
printf("Hello, I'm %s years old.", 22);
Unfortunately in C there is no way to avoid these relatively common mistakes.
gets
The gets function is used for a completely different purpose: it reads a string from the standard input.
For example:
char name[512];
printf("What's your name? ");
gets(name);
This simple program will ask the user for a name and save what he or she types into name.
However, gets() should NEVER be used. It will open your application and the system it runs on to security vulnerabilities.
Quoting from the man page:
Never use gets(). Because it is
impossible to tell without knowing the
data in advance how many characters
gets() will read, and because gets()
will continue to store characters past
the end of the buffer, it is extremely
dangerous to use. It has been used to
break computer security. Use fgets()
instead.
Explained in a more simple way the problem is that if the variable you give gets (name in this case) is not big enough to hold what the user types a buffer overflow will occur, which is, gets will write past the end of the variable. This is undefined behavior and on some systems it will allow execution of arbitrary code by the attacker.
Since the variable must have a finite, static size and you can't set a limit of the amount of characters the user can type as the input, gets() is never secure and should never be used. It exists only for historical reasons.
As the manual suggested, you should use fgets instead. It has the same purpose as gets but has a size argument that specifies the size of the variable:
char *fgets(char *s, int size, FILE *stream);
So, the program above would become:
char name[512];
printf("What's your name? ");
fgets(name, sizeof(name) /* 512 */, stdin /* The standard input */);
They fundamentally perform different tasks.
printf: prints out text to a console.
gets: reads in input from the keyboard.
printf: allowing you to format a string from components (ie. taking results from variables), and when output to stdout, it does not append new line character. You have to do this by inserting '\n' in the format string.
puts: only output a string to stdout, but does append new line afterward.
scanf: scan the input fields, one character at a time, and convert them according to the given format.
gets: simply read a string from stdin, with no format consideration, the return character is replaced by string terminator '\0'.
http://en.wikipedia.org/wiki/Printf
http://en.wikipedia.org/wiki/Gets

Disadvantages of scanf

I want to know the disadvantages of scanf().
In many sites, I have read that using scanf might cause buffer overflows. What is the reason for this? Are there any other drawbacks with scanf?
Most of the answers so far seem to focus on the string buffer overflow issue. In reality, the format specifiers that can be used with scanf functions support explicit field width setting, which limit the maximum size of the input and prevent buffer overflow. This renders the popular accusations of string-buffer overflow dangers present in scanf virtually baseless. Claiming that scanf is somehow analogous to gets in the respect is completely incorrect. There's a major qualitative difference between scanf and gets: scanf does provide the user with string-buffer-overflow-preventing features, while gets doesn't.
One can argue that these scanf features are difficult to use, since the field width has to be embedded into format string (there's no way to pass it through a variadic argument, as it can be done in printf). That is actually true. scanf is indeed rather poorly designed in that regard. But nevertheless any claims that scanf is somehow hopelessly broken with regard to string-buffer-overflow safety are completely bogus and usually made by lazy programmers.
The real problem with scanf has a completely different nature, even though it is also about overflow. When scanf function is used for converting decimal representations of numbers into values of arithmetic types, it provides no protection from arithmetic overflow. If overflow happens, scanf produces undefined behavior. For this reason, the only proper way to perform the conversion in C standard library is functions from strto... family.
So, to summarize the above, the problem with scanf is that it is difficult (albeit possible) to use properly and safely with string buffers. And it is impossible to use safely for arithmetic input. The latter is the real problem. The former is just an inconvenience.
P.S. The above in intended to be about the entire family of scanf functions (including also fscanf and sscanf). With scanf specifically, the obvious issue is that the very idea of using a strictly-formatted function for reading potentially interactive input is rather questionable.
The problems with scanf are (at a minimum):
using %s to get a string from the user, which leads to the possibility that the string may be longer than your buffer, causing overflow.
the possibility of a failed scan leaving your file pointer in an indeterminate location.
I very much prefer using fgets to read whole lines in so that you can limit the amount of data read. If you've got a 1K buffer, and you read a line into it with fgets you can tell if the line was too long by the fact there's no terminating newline character (last line of a file without a newline notwithstanding).
Then you can complain to the user, or allocate more space for the rest of the line (continuously if necessary until you have enough space). In either case, there's no risk of buffer overflow.
Once you've read the line in, you know that you're positioned at the next line so there's no problem there. You can then sscanf your string to your heart's content without having to save and restore the file pointer for re-reading.
Here's a snippet of code which I frequently use to ensure no buffer overflow when asking the user for information.
It could be easily adjusted to use a file other than standard input if necessary and you could also have it allocate its own buffer (and keep increasing it until it's big enough) before giving that back to the caller (although the caller would then be responsible for freeing it, of course).
#include <stdio.h>
#include <string.h>
#define OK 0
#define NO_INPUT 1
#define TOO_LONG 2
#define SMALL_BUFF 3
static int getLine (char *prmpt, char *buff, size_t sz) {
int ch, extra;
// Size zero or one cannot store enough, so don't even
// try - we need space for at least newline and terminator.
if (sz < 2)
return SMALL_BUFF;
// Output prompt.
if (prmpt != NULL) {
printf ("%s", prmpt);
fflush (stdout);
}
// Get line with buffer overrun protection.
if (fgets (buff, sz, stdin) == NULL)
return NO_INPUT;
// Catch possibility of `\0` in the input stream.
size_t len = strlen(buff);
if (len < 1)
return NO_INPUT;
// If it was too long, there'll be no newline. In that case, we flush
// to end of line so that excess doesn't affect the next call.
if (buff[len - 1] != '\n') {
extra = 0;
while (((ch = getchar()) != '\n') && (ch != EOF))
extra = 1;
return (extra == 1) ? TOO_LONG : OK;
}
// Otherwise remove newline and give string back to caller.
buff[len - 1] = '\0';
return OK;
}
And, a test driver for it:
// Test program for getLine().
int main (void) {
int rc;
char buff[10];
rc = getLine ("Enter string> ", buff, sizeof(buff));
if (rc == NO_INPUT) {
// Extra NL since my system doesn't output that on EOF.
printf ("\nNo input\n");
return 1;
}
if (rc == TOO_LONG) {
printf ("Input too long [%s]\n", buff);
return 1;
}
printf ("OK [%s]\n", buff);
return 0;
}
Finally, a test run to show it in action:
$ printf "\0" | ./tstprg # Singular NUL in input stream.
Enter string>
No input
$ ./tstprg < /dev/null # EOF in input stream.
Enter string>
No input
$ ./tstprg # A one-character string.
Enter string> a
OK [a]
$ ./tstprg # Longer string but still able to fit.
Enter string> hello
OK [hello]
$ ./tstprg # Too long for buffer.
Enter string> hello there
Input too long [hello the]
$ ./tstprg # Test limit of buffer.
Enter string> 123456789
OK [123456789]
$ ./tstprg # Test just over limit.
Enter string> 1234567890
Input too long [123456789]
From the comp.lang.c FAQ: Why does everyone say not to use scanf? What should I use instead?
scanf has a number of problems—see questions 12.17, 12.18a, and 12.19. Also, its %s format has the same problem that gets() has (see question 12.23)—it’s hard to guarantee that the receiving buffer won’t overflow. [footnote]
More generally, scanf is designed for relatively structured, formatted input (its name is in fact derived from “scan formatted”). If you pay attention, it will tell you whether it succeeded or failed, but it can tell you only approximately where it failed, and not at all how or why. You have very little opportunity to do any error recovery.
Yet interactive user input is the least structured input there is. A well-designed user interface will allow for the possibility of the user typing just about anything—not just letters or punctuation when digits were expected, but also more or fewer characters than were expected, or no characters at all (i.e., just the RETURN key), or premature EOF, or anything. It’s nearly impossible to deal gracefully with all of these potential problems when using scanf; it’s far easier to read entire lines (with fgets or the like), then interpret them, either using sscanf or some other techniques. (Functions like strtol, strtok, and atoi are often useful; see also questions 12.16 and 13.6.) If you do use any scanf variant, be sure to check the return value to make sure that the expected number of items were found. Also, if you use %s, be sure to guard against buffer overflow.
Note, by the way, that criticisms of scanf are not necessarily indictments of fscanf and sscanf. scanf reads from stdin, which is usually an interactive keyboard and is therefore the least constrained, leading to the most problems. When a data file has a known format, on the other hand, it may be appropriate to read it with fscanf. It’s perfectly appropriate to parse strings with sscanf (as long as the return value is checked), because it’s so easy to regain control, restart the scan, discard the input if it didn’t match, etc.
Additional links:
longer explanation by Chris Torek
longer explanation by yours truly
References: K&R2 Sec. 7.4 p. 159
It is very hard to get scanf to do the thing you want. Sure, you can, but things like scanf("%s", buf); are as dangerous as gets(buf);, as everyone has said.
As an example, what paxdiablo is doing in his function to read can be done with something like:
scanf("%10[^\n]%*[^\n]", buf));
getchar();
The above will read a line, store the first 10 non-newline characters in buf, and then discard everything till (and including) a newline. So, paxdiablo's function could be written using scanf the following way:
#include <stdio.h>
enum read_status {
OK,
NO_INPUT,
TOO_LONG
};
static int get_line(const char *prompt, char *buf, size_t sz)
{
char fmt[40];
int i;
int nscanned;
printf("%s", prompt);
fflush(stdout);
sprintf(fmt, "%%%zu[^\n]%%*[^\n]%%n", sz-1);
/* read at most sz-1 characters on, discarding the rest */
i = scanf(fmt, buf, &nscanned);
if (i > 0) {
getchar();
if (nscanned >= sz) {
return TOO_LONG;
} else {
return OK;
}
} else {
return NO_INPUT;
}
}
int main(void)
{
char buf[10+1];
int rc;
while ((rc = get_line("Enter string> ", buf, sizeof buf)) != NO_INPUT) {
if (rc == TOO_LONG) {
printf("Input too long: ");
}
printf("->%s<-\n", buf);
}
return 0;
}
One of the other problems with scanf is its behavior in case of overflow. For example, when reading an int:
int i;
scanf("%d", &i);
the above cannot be used safely in case of an overflow. Even for the first case, reading a string is much more simpler to do with fgets rather than with scanf.
Yes, you are right. There is a major security flaw in scanf family(scanf,sscanf, fscanf..etc) esp when reading a string, because they don't take the length of the buffer (into which they are reading) into account.
Example:
char buf[3];
sscanf("abcdef","%s",buf);
clearly the the buffer buf can hold MAX 3 char. But the sscanf will try to put "abcdef" into it causing buffer overflow.
Problems I have with the *scanf() family:
Potential for buffer overflow with %s and %[ conversion specifiers. Yes, you can specify a maximum field width, but unlike with printf(), you can't make it an argument in the scanf() call; it must be hardcoded in the conversion specifier.
Potential for arithmetic overflow with %d, %i, etc.
Limited ability to detect and reject badly formed input. For example, "12w4" is not a valid integer, but scanf("%d", &value); will successfully convert and assign 12 to value, leaving the "w4" stuck in the input stream to foul up a future read. Ideally the entire input string should be rejected, but scanf() doesn't give you an easy mechanism to do that.
If you know your input is always going to be well-formed with fixed-length strings and numerical values that don't flirt with overflow, then scanf() is a great tool. If you're dealing with interactive input or input that isn't guaranteed to be well-formed, then use something else.
Many answers here discuss the potential overflow issues of using scanf("%s", buf), but the latest POSIX specification more-or-less resolves this issue by providing an m assignment-allocation character that can be used in format specifiers for c, s, and [ formats. This will allow scanf to allocate as much memory as necessary with malloc (so it must be freed later with free).
An example of its use:
char *buf;
scanf("%ms", &buf); // with 'm', scanf expects a pointer to pointer to char.
// use buf
free(buf);
See here. Disadvantages to this approach is that it is a relatively recent addition to the POSIX specification and it is not specified in the C specification at all, so it remains rather unportable for now.
The advantage of scanf is once you learn how use the tool, as you should always do in C, it has immensely useful usecases. You can learn how to use scanf and friends by reading and understanding the manual. If you can't get through that manual without serious comprehension issues, this would probably indicate that you don't know C very well.
scanf and friends suffered from unfortunate design choices that rendered it difficult (and occasionally impossible) to use correctly without reading the documentation, as other answers have shown. This occurs throughout C, unfortunately, so if I were to advise against using scanf then I would probably advise against using C.
One of the biggest disadvantages seems to be purely the reputation it's earned amongst the uninitiated; as with many useful features of C we should be well informed before we use it. The key is to realise that as with the rest of C, it seems succinct and idiomatic, but that can be subtly misleading. This is pervasive in C; it's easy for beginners to write code that they think makes sense and might even work for them initially, but doesn't make sense and can fail catastrophically.
For example, the uninitiated commonly expect that the %s delegate would cause a line to be read, and while that might seem intuitive it isn't necessarily true. It's more appropriate to describe the field read as a word. Reading the manual is strongly advised for every function.
What would any response to this question be without mentioning its lack of safety and risk of buffer overflows? As we've already covered, C isn't a safe language, and will allow us to cut corners, possibly to apply an optimisation at the expense of correctness or more likely because we're lazy programmers. Thus, when we know the system will never receive a string larger than a fixed number of bytes, we're given the ability to declare an array that size and forego bounds checking. I don't really see this as a down-fall; it's an option. Again, reading the manual is strongly advised and would reveal this option to us.
Lazy programmers aren't the only ones stung by scanf. It's not uncommon to see people trying to read float or double values using %d, for example. They're usually mistaken in believing that the implementation will perform some kind of conversion behind the scenes, which would make sense because similar conversions happen throughout the rest of the language, but that's not the case here. As I said earlier, scanf and friends (and indeed the rest of C) are deceptive; they seem succinct and idiomatic but they aren't.
Inexperienced programmers aren't forced to consider the success of the operation. Suppose the user enters something entirely non-numeric when we've told scanf to read and convert a sequence of decimal digits using %d. The only way we can intercept such erroneous data is to check the return value, and how often do we bother checking the return value?
Much like fgets, when scanf and friends fail to read what they're told to read, the stream will be left in an unusual state;
In the case of fgets, if there isn't sufficient space to store a complete line, then the remainder of the line left unread might be erroneously treated as though it's a new line when it isn't.
In the case of scanf and friends, a conversion failed as documented above, the erroneous data is left unread on the stream and might be erroneously treated as though it's part of a different field.
It's no easier to use scanf and friends than to use fgets. If we check for success by looking for a '\n' when we're using fgets or by inspecting the return value when we use scanf and friends, and we find that we've read an incomplete line using fgets or failed to read a field using scanf, then we're faced with the same reality: We're likely to discard input (usually up until and including the next newline)! Yuuuuuuck!
Unfortunately, scanf both simultaneously makes it hard (non-intuitive) and easy (fewest keystrokes) to discard input in this way. Faced with this reality of discarding user input, some have tried scanf("%*[^\n]%*c");, not realising that the %*[^\n] delegate will fail when it encounters nothing but a newline, and hence the newline will still be left on the stream.
A slight adaptation, by separating the two format delegates and we see some success here: scanf("%*[^\n]"); getchar();. Try doing that with so few keystrokes using some other tool ;)
There is one big problem with scanf-like functions - the lack of any type safety. That is, you can code this:
int i;
scanf("%10s", &i);
Hell, even this is "fine":
scanf("%10s", i);
It's worse than printf-like functions, because scanf expects a pointer, so crashes are more likely.
Sure, there are some format-specifier checkers out there, but, those are not perfect and well, they are not part of the language or the standard library.

Resources