Question

I’m a a beginner to C.

main() {    
   int *a-ptr = (int *)malloc(int); 
   *a-ptr = 5;
   printf(“%d”, *a-ptr);
}

The question is: is this guaranteed to print 5?

The answer is : NO, for two reasons:

• you can’t use “-” in the names of variables
• “you didn’t allocate int storage”

I don’t understand that second point. Isn’t the storage allocated with this line?

int *a-ptr = (int *)malloc(int);

Answer 1

    main() {    // wrong. Should return int
int main() {    // better

int *a-ptr =  //wrong. no dashes in variable names
int *a_ptr =  // better, use underscores if you want to have multiparted names

(int *)malloc(int); // wrong. Don't typecast the return of malloc(), also it takes a 
                    // size, not a type
malloc(sizeof(int)); // better, you want enought memory for the sizeof 1 int 

So a better version of your code would be:

int main() {    
   int *a_ptr = malloc(sizeof(int)); 
   *a_ptr = 5;
   printf("%d", *a_ptr);
   free(a_ptr); // When you're done using memory allocated with malloc, free it
   return 0;
}