I'm trying to understand malloc but I keep getting "Segmentation fault: 11" with this piece of code:
#include <stdio.h> #include <stdlib.h> int main() { int i = 0, j = 0; char ** ptr = (char **) malloc(sizeof(char*)); for(i = 0; i < 5; i++) { for(j = 0; j < 10; j++) ptr[i][j] = 'a'; printf("%s\n", ptr[i]); } return 0; } I thought there wasn't enough bytes being allocated so I did malloc(sizeof(char*) * 100, but gives me the same error. What am I not understanding here?
When you allocate a 2D array, you need to allocate the individual sub-arrays as well. In addition, you need to say how many elements you wish to have. For that you multiply the desired count by the number of elements, like this:
char ** ptr = (char **) malloc(5*sizeof(char*)); // Size=5 ---------------------^ for(int i = 0; i < 5; i++) { ptr[i] = malloc(11*sizeof(char)); // sizeof(char) is always 1, so the multiplication above is redundant. // You need 11 elements for ten characters for(int j = 0; j < 10; j++) { ptr[i][j] = 'a'; } // don't forget to null-terminate the string: ptr[i][10] = '\0'; printf("%s\n", ptr[i]); } 
Your code is totally messed up in every aspect!
1) you allocated memory for exactly 1 Pointer to it. This means you can access ptr[0], but not ptr[1] ... ptr[4] as you are trying to do.
2) you never allocate anything for the elements in ptr[i].
3) you try to print a string a ptr[i] which is (even if your allocation would be right) never terminated.
4) although this is obviously only a beginners test, never forget to free your memory!!!!
To reach something CLOSE to what your sampel code is describing you could do:
int main() { int i,j; char ** ptr = malloc( 5 * sizeof(char*) ); /* an array of 5 elements of type char* */ for(i = 0; i < 5; i++) { ptr[i] = malloc( 11*sizeof(char) ); /* The element i of the array is an array of 11 chars (10 for the 'a' character, one for the null-termination */ for(j = 0; j < 10; j++) ptr[i][j] = 'a'; ptr[i][10] = '\0'; /* strings need to be null terminated */ printf("%s\n", ptr[i]); } // free your memory! for (i=0; i<5; i++ ) { free(ptr[i]); } free(ptr); return 0; 
free(). I should use it once I'm done using any pointer initialized with malloc?
ptr[0] either as he has not initialized it. He allocated memory for lots of pointers but did not make them point anywhere.Another way to allocate the memory is:
char (*ptr)[11] = malloc( sizeof(char[5][11]) ); for(i = 0; i < 5; i++) { for(j = 0; j < 10; j++) ptr[i][j] = 'a'; ptr[i][10] = 0; printf("%s\n", ptr[i]); } free(ptr); It seems less hassle to use a single allocation than to use a lot of allocations, unless you have a pressing reason to do the latter.
char *ptr[11];
ptr[i]is a pointer, but what does it point to ?char * ptr = (char *) malloc(sizeof(char))and take away the second loop statement, it works fine.ptrdoes, butptr[0]does not.malloc& friends in C!