scanf() not working properly?

Question

  char *str;
  printf("Enter string:\n");
  scanf("%s",str);

OUTPUT: runtime-check failure#3 str is being used without being initialized

You are trying to read data into memory location pointed by an uninitialized pointer? — AnT stands with Russia, Sep 28 '13 at 16:34
It is odd that everyone assumes 'need a string; use `malloc()`' rather than 'need a string; declare an array'. The array is easier to manage; no-one has reminded you of the need to `free()` that which is allocated via `malloc()`, nor of the need to check the return value from `malloc()`. — Jonathan Leffler, Sep 28 '13 at 16:48

score 3 · Answer 1 · answered Sep 28 '13 at 16:42

Allocate an array and read into that:

char str[100];

if (scanf("%99s", str) != 1)
    ...error...

Or, if you need a pointer, then:

char data[100];
char *str = data;

if (scanf("%99s", str) != 1)
    ...error...

Note the use of a length to prevent buffer overflow. Note that the length specified to scanf() et al is one less than the total length (an oddity based on ancient precedent; most code includes the null byte in the specified length — see fgets(), for example).

Remember that %s will skip past leading white space and then stop on the first white space after some non-white space character. In particular, it will leave the newline in the input stream, ready for the next input operation to read. If you want the whole line of input, then you should probably use fgets() and sscanf() rather than raw scanf() — in fact, very often you should use fgets() and sscanf() rather than scanf() or fscanf(), if only because it make sensible error reporting a lot easier.

Rahul Tripathi · Answer 2 · 2013-09-28T16:37:33.440

1

Its an undefined behavior if you dont initialize it.You have an uninitialized pointer which is reading data into memory location which may eventually cause trouble for you. You've declared str as a pointer, but you haven't given it a valid location to point to; it initially contains some random value that may or may not be a writable memory address.Try to allocate memory to the char *str;

char *str = malloc(sizeof(char)*100);

edited Sep 28 '13 at 16:37

answered Sep 28 '13 at 16:28

Rahul Tripathi

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memory allocation just to have user input :-/, R.I.P C language – hina abbasi Sep 28 '13 at 16:33
Without having enough space in a BOX. How can you able to put some thing extra things in that.allocating memory is nothing but creating some space – Gangadhar Sep 28 '13 at 16:41
Why on EARTH would you `malloc` this instead of declaring an array? Is this specifically so you can forget to `free` it later? Please don't do this! – Dawood ibn Kareem Apr 10 '14 at 10:04

score 0 · Answer 3 · answered Sep 28 '13 at 16:28

0

char *str declares str as a pointer to char type. scanf("%s",str) will read only E from the entered string.

answered Sep 28 '13 at 16:28

haccks

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score 0 · Answer 4 · edited May 23 '17 at 12:03

0

You might NOT want to use exactly scanf("%s") (or get in the habit of using it) as it is vulnerable to buffer overflow (i.e. might accept more characters than your buffer can hold). For a discussion on this see Disadvantages of scanf

edited May 23 '17 at 12:03

Community

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answered Sep 28 '13 at 16:39

halfbit

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score 0 · Answer 5 · answered Sep 28 '13 at 16:46

You have to allocate memory before assigning some value to a pointer. Always remember that pointer points to a memory location and you have to tell him what memory location you want to assign for him. So, for doing that you have to do :

str = (char*)malloc(sizeof(char));

This will assign you 1byte of memory block. So, you can assign only one character in this memory block. For a string you have to assign as many blocks depending on the number of characters in the string.Say, "Stack" requires 5 character and 1 extra block for '\0'. So, you have to assign at least 6 memory block to him.

str =  (char*)malloc(6*sizeof(char));

I hope this will help you to grow more concepts in C.

You shouldn't describe how to use `malloc` without also describing how to use `free`, and explaining why one must always accompany the other. Beginners reading your answer are going to go and create their very own memory leaks if you don't. Really, in this case, using an array is a much better solution. — Dawood ibn Kareem, Apr 10 '14 at 10:07

scanf() not working properly?

5 Answers5