Malloc and scanf

Question

I\'m fairly competent in a few scripting languages, but I\'m finally forcing myself to learn raw C. I\'m just playing around with some basic stuff (I/O right now). How can

Answer 1

  char *toParseStr = (char*)malloc(10);
  printf("Enter string here: ");
  scanf("%s",toParseStr);
  printf("%s",toParseStr);
  free(toParseStr);

Firstly, the string in scanf is specifies the input it's going to receive. In order to display a string before accepting keyboard input, use printf as shown.

Secondly, you don't need to dereference toParseStr since it's pointing to a character array of size 10 as you allocated with malloc. If you were using a function which would point it to another memory location, then &toParseStr is required.

For example, suppose you wanted to write a function to allocate memory. Then you'd need &toParseStr since you're changing the contents of the pointer variable (which is an address in memory --- you can see for yourself by printing its contents).

void AllocateString(char ** ptr_string, const int n)
{
    *ptr_string = (char*)malloc(sizeof(char) * n);
}

As you can see, it accepts char ** ptr_string which reads as a pointer which stores the memory location of a pointer which will store the memory address (after the malloc operation) of the first byte of an allocated block of n bytes (right now it has some garbage memory address since it is uninitialized).

int main(int argc, char *argv[])
{
  char *toParseStr;
  const int n = 10;
  printf("Garbage: %p\n",toParseStr);
  AllocateString(&toParseStr,n);
  printf("Address of the first element of a contiguous array of %d bytes: %p\n",n,toParseStr);

  printf("Enter string here: ");
  scanf("%s",toParseStr);
  printf("%s\n",toParseStr);
  free(toParseStr);

  return 0;
}

Thirdly, it is recommended to free memory you allocate. Even though this is your whole program, and this memory will be deallocated when the program quits, it's still good practice.