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Home/ Questions/Q 939987
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Editorial Team
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Editorial Team
Asked: May 15, 20262026-05-15T21:52:51+00:00 2026-05-15T21:52:51+00:00

I’m fairly competent in a few scripting languages, but I’m finally forcing myself to

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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 I allocate heap memory, store a string in the allocated memory, and then spit it back out out? This is what I have right now, how can I make it work correctly?

#include <stdio.h>
#include <stdlib.h>

int main(int argc, char *argv[])
{
  char *toParseStr = (char*)malloc(10);
  scanf("Enter a string",&toParseStr);
  printf("%s",toParseStr);
  return 0;
}

Currently I’m getting weird output like ‘8’\’.

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  1. Editorial Team
    Editorial Team
    2026-05-15T21:52:51+00:00Added an answer on May 15, 2026 at 9:52 pm
      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.

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