If I have
What have I created? What's array pointing to?Code:char **array = malloc (sizeof *array);
If I write
How many array[row] = malloc (size); can I write, given that row is increased each time? Is it correct to do that?Code:size_t row; /* defined before calling malloc */ size_t size; /* defined later before calling malloc */ . . . array[row] = malloc (size); . . . ++row;
A pointer to a pointer. Of whatever type "array" belongs.Originally Posted by H`eya
This looks like something you might use this way:How many array[row] = malloc (size); can I write, given that row is increased each time? Is it correct to do that?
in which case you would have an array of ten pointers to arrays of SIZE ints.Code:int *array[10]; for (row=0;row<10;row++) array[row] = malloc(SIZE*sizeof(int))
C programming resources:
GNU C Function and Macro Index -- glibc reference manual
The C Book -- nice online learner guide
Current ISO draft standard
CCAN -- new CPAN like open source library repository
3 (different) GNU debugger tutorials: #1 -- #2 -- #3
cpwiki -- our wiki on sourceforge
Appreciate your prompt response.
Allow me to explain further then, since I think I did not give you enough to clarify my concerns.
I apologize that I did not explain well in my previous post.Code:#include <stdio.h> #include <stdlib.h> int main (void ) { size_t i; size_t row = 5, size = 32; char **array = malloc(sizeof *array); if (!array) { /* handle the error */ } /* * This is the part I am interested in * knowing if it would be correct */ for (i = 0, i < row, i++) { *array[i] = malloc(size); /* does **array have memory for *array[i]? */ if (*array == NULL) { /* handle the error */ } } return 0; }
malloc gives you the amount of memory asked for. You ask for enough memory to hold one of "* array" things. So you have enough room for one, not 5.
Many thanks, to both of you.
I understand.
