Can't pass pointer to strcat function

Question

I'm trying to concatenate a string to another string using strcat function available in string.h, it compiles successfully but on execution, it gives segmentation fault (core dumped) message on terminal. This is my code:

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

 int main()
 {
   char* dest;
   char* src;

   dest = malloc(16);
   src = malloc(11);

   dest = "text/";
   src = "javascript";

   strcat(dest, src);

   printf("%s\n", dest);

   free(dest);
   free(src);
   return 0;
}

please tell me where is memory infinitely used for which it is giving segmentation fault.

Answer 1

It is related to how you are storing the strings in src and dest. When you store them as you did, they end up concatenated in storage. Also, this doesn't store the \0 end of string marker. It also ends up putting src at dest+5. I'm not clear why this happens, but it causes a lot of grief.

Now, if you were to do this instead:

 strcpy(src,  "javascript");
 strcpy(dest, "text/");

it will work as intended. strcpy takes care of all the housekeeping, including storing the \0 end of string marker. When you do the strcat later, it works as intended.

If this answers your question, please accept this answer to remove the question from the unanswered question pool. Let's keep up on forum housekeeping. ;-)

[edit] For the why of why it didn't work, see @kareem's explanation below, particularly the part after "but when you do that". Nice job, @kareem.

Answer 2

you need to understand the function of the assignment operator (i.e., =). think of what happens when you do something like

int i = GetInt();
i = 20;

after assigning 20 to i, can we get whatever was in i before the assignment (the value returned by GetInt) back?

well, unless we stored it in another variable or something, apparently we can't. that's because, assuming the assignment went well (i.e., no errors, etc), assigning a value to a variable overwrites the previous value (whatever was in there before the assignment) forever.

now when you do

char* dest;
char* src;

dest = malloc(16);
src = malloc(11);

no problem with that. you're obviously defining two pointers, allocating memory for them on the heap by calling malloc which according to its man page returns the address of the allocated memory (if the allocation was successful). the returned addresses are stored in the pointers normally.

but when you then do that

dest = "text/";
src = "javascript";

you're not storing "text/" and "javascript" into the memory pointed to by dest and src respectively. rather you're losing the addresses that were in there forever and having the pointers pointing to these strings instead.

string literals in C are compile-time constants. they are stored in a read-only data segment in memory and they cannot be modified. so now the pointers are pointing to locations on a read-only memory.

the call to strcat(dest, src) tries to modify the memory pointed to by dest (which recall should not be modified) so you get a segfault.

---

the fix:

you may use a function like strcpy from the standard library to initialize the memory pointed to by the pointers after allocating memory for them. execute man strcpy for more info!