Thread: Question on malloc() and reading strings

Hi,

I've just switched from java programming to c programming and am currently facing some problems (probably as I'm unfamilar with pointers and c syntax).

Let's same my program has to read in an unknown number of strings (with 40 chars max) until it reads the string "halt". I have to use the malloc function to ensure that the correct amount of memory is allocated to the input string, and then print out the size of this new string (as proof).

Why is it that the size of the new string is fixed at the size of the very first string processed? Guess that I'm doing something seriously wrong here.

Code:

int main() {

   char[41] currString;
   char *newString;
   scanf("%s",currString);
   while (strcmp(currString,"stop")!=0) {
newString=(char *) malloc (strlen(currString)+1);
printf("%lu\n",sizeof(newString));
scanf("%s",currString);
   return 0;

}

[/code]

Hope to get some advice. Thanks.

Originally Posted by rockysfr

Code:

int main() {

   char currString[41];
   char *newString;
   scanf("%s",currString);
   while (strcmp(currString,"stop")!=0) {
newString=(char *) malloc (strlen(currString)+1);
printf("%lu\n",sizeof(newString));
scanf("%s",currString);
   return 0;

}

The sizeof operator returns the size of the object, not what the object contains. So sizeof(any_pointer) on 32-bit compilers is always 4, regardless of how many bytes is allocated for the pointer. There is no known method to later get the number of bytes allocated for the pointer, your program must remember that information.

Your program has another bug -- memory allocated by malloc() must be released by free(). If you don't do that, your program will "leak" memory, meaning that the memory will be lost and cannot be recovered until the program terminated.