How to read a string into a void pointer?

Question

I have been trying to read an input as string from the user inside a void pointer in C. SO i wrote something like the following:

void *ptr;
ptr = calloc(100,sizeof(char));
printf("Enter the string: ");
fgets(*((char *)ptr),100,stdin);
printf("You entered ");
puts(*((char *)ptr));

I know I may not be doing it the right way, so can anybody please help me show the right way of taking a string input in a void pointer? I want something as

input:- Enter the string: welcome user

output:- You entered: welcome user

Answer 1

Just convert the void* to a char*:

void *ptr;
ptr = calloc(100,sizeof(char));
printf("Enter the string: ");
fgets((char*)ptr,100,stdin);
printf("You entered ");
puts((char*)ptr);

fgets and puts take a pointer as first argument, so you could use (char*)ptr to convert the pointer.

If you write *((char*)ptr) you treat the void pointer as a char pointer, but also dereference it with * which will give you the first character. This is not what you want here.

Answer 2

You need to remove the (char *) casts. Try this:

void *ptr;
ptr = calloc(100,sizeof(char));
printf("Enter the string: ");
fgets((char *)ptr,100,stdin);
printf("You entered ");
puts((char *)ptr);

Answer 3

You are dereferencing the pointer when you pass it to fgets and fputs which means you pass the value of the first character in the memory location. Since you used calloc for the allocation you pass '\0' (the null character, zero).

Your ptr (and also (char *)ptr) points to some memory, i.e. its value is an address, a point in the RAM of your computer. fgets will copy chars from the console input to that memory location, byte by byte. Your expression *((char *)ptr) which you pass as parameter to fgets is, by contrast, the value stored at that location. After calloc() that value is zero, i.e. the character '\0'. Fgets thinks that's a pointer, tries to access the memory at address 0, and crashes. The same would happen with fputs, because under modern PC operating systems even a read access at "weird" addresses is not allowed for user programs. The solution is, as the others pointed out correctly, to omit the dereferencing "*".

By the way, did you not get a compiler warning? While C is not really type safe, it does have function prototypes, and most compilers warn against wrong parameter types when they are called.

Answer 4

What you basically want cannot be achieved due to your arguments in fgets().

As the function fgets() take three arguments

1.pointer to the first cell of memory at where to store the data.
2.No of characters to be written to memeory.
3.stream pointer,from where to read.

But what you are doing wrong is with first argument,as you are not passing a pointer to fgets().You are passing *((char *)ptr) it is of type char because of your unnecessary derefferencing.Removing external * will cure it because its type now becomes char * and you can use it as a legal argument for fgets().