How to check if a string returned by scanf is null

Question

I'm reading an input string from the user with scanf().

I want to check if this string is NULL (\0) or not.

Here is my code:

#include<stdio.h>

char *argument; // argument for mycat
scanf("%s", &argument);

if(fork()==0)       // at child
{
    printf("Child process: about to execute \"mycat %s\"\n", &argument);
    fflush(stdout);
    if(strcmp(argument, "") == 0) // <-- Here is the problem
    {

        execlp("mycat", "mycat", &argument, NULL);      // execute child process
    }
    execlp("mycat","mycat", NULL);
}

I'm compiling with g++ compiler on Red Hat 6.1

Edit: The problem is that I'm not able to dereference argument either for the if statement or even for use with strlen().

why not use strlen instead, if the length is zero then it would work right? — osirisgothra, Mar 02 '14 at 11:16
without entering text if you press ENTER it will not return the null.. it return the empty string mean length is zero.. so i will also suggest you to check if **strlen** is 0 mean user didn't enter any value. — Deepak Sharma, Mar 02 '14 at 11:19
`char *argument;` argument is uninitialised; it does not point to anything. — wildplasser, Mar 02 '14 at 11:35
Alright, apparently my question is written in a wrong way. I was thinking that when the user inputs `"\0"` it will be equivalent to a null string, however `strlen()` returns a length of 2. So I finally used `strcmp(input, "\\0")` to check if user input is equal to `"\0"` and it worked. I'm accepting the first answer because it's more relevant to the question and it's the first one. I apologize for the confusion! — Mahozi, Mar 02 '14 at 17:47

abligh · Accepted Answer · 2014-03-02T12:05:20.727

4

NULL and \0 are not the same thing, though they both evaluate to zero. NULL is a pointer zero, i.e. it's what we use for a null pointer. \0 is the character with ASCII number zero, otherwise known as NUL (one 'L'), i.e. is a char with value 0.

There is an important difference between a char * being NULL (i.e. there not being a string in memory at all), or the string being empty (i.e. only containing one char which is \0 otherwise known as NUL).

To test for the first:

if (!string)

or if you like to be more verbose:

if (string == NULL)

To test for the second:

if (!string[0])

or if you like to be more verbose:

if (string[0] == 0)

Obviously if you need to test for both, test for the first then the second, as the second would dereference a null pointer if string was NULL.

edited Mar 02 '14 at 12:05

answered Mar 02 '14 at 11:22

abligh

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Thanks for clarifying the difference, the second situation is what I want. However, this solution generates a run-time error and the program never enters the 'if' statement. – Mahozi Mar 02 '14 at 14:46
@Ameen There are other things wrong with your program, primarily that you have not allocated any memory for your string. Try `char argument[1024];`. – abligh Mar 02 '14 at 16:21

alk · Answer 2 · 2014-03-02T17:57:49.780

0

To answer the question:

Check scanf()'s return value to test whether the operation succeeded, read in something.

For your example it would return 0 if nothing was read, or EOF if there was nothing to read from or an error occured.

However there are issues with the example code:

1^st issue:

char *argument; // argument for mycat

is just a pointer pointing to "somewhere/nowhere". Reading an uninitalised variable invokes undefined behaviour, copying data to "nowhere" also invokes undefined behaviour.

To fix this allocate some memory to it by for example doing:

char * argument = malloc(42); /* Allocate 41+1 bytes, that is room for a C-"string" 
                                 being 41 characters long. */
strcpy(argument, ""); /* Initalise it to an emtpy string. */

2^nd issue:

scanf("%s", &argument);

tries to scan data to the address of argument not where it is pointing to (if it were).

To fix this do

scanf("%s", argument);

or even better, also tell scanf() the size if the buffer to read into, so it won't overflow it. SO assuming the above initialisation, this would be:

scanf("%41s", argument); /* Read one less then there is room, as the 42nd char is used to hold the '0'-terminator. */

edited Mar 02 '14 at 17:57

answered Mar 02 '14 at 11:48

alk

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`char * argument = malloc(42);` gives me a compilation error: `Invalid conversion from void* to char*`. – Mahozi Mar 02 '14 at 14:40
@Ameen: This error indicates you are actually not using a C compiler, but a C++ compiler. – alk Mar 02 '14 at 14:44
@Ameen you tagged this "C" where casting the return from `malloc` is harmless but frowned upon. In C++ you need to cast the returned value of malloc; I am told the preferred way in C++ to do that is to use `static_cast<>`, per http://stackoverflow.com/questions/103512/in-c-why-use-static-castintx-instead-of-intx – abligh Mar 02 '14 at 16:20
alk&@abligh You're right, I was compiling this with g++, it worked fine with gcc thou. I used `strcmp()` to check user input. – Mahozi Mar 02 '14 at 17:40

How to check if a string returned by scanf is null

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