I am lost. I cannot assign an int
to a dereferenced int *
.
printf("in octave\n");
int *default_octave;
printf("attr[%d]: %s\n",i+1,attr[i+1]);
const char *octave_char = attr[i+1];
printf("octave_char: %s\n", octave_char);
int octave_number = atoi(octave_char);
printf("octave_number: %d\n", octave_number);
fflush(stdout);
*default_octave=octave_number;
printf("in octave pt 2\n");
fflush(stdout);
This is the output:
in octave
attr[1]: 4
octave_char: 4
octave_number: 4
Segmentation fault
Why?
Running the GDB debugger gets to that line and then seg faults, too.
4
0 int octave_number = atoi(octave_char);
(gdb) s
41 printf("octave_number: %d\n", octave_number);
(gdb)
octave_number: 4
42 fflush(stdout);
(gdb)
43 *default_octave=octave_number;
(gdb) print octave_number
$1 = 4
(gdb) s
Program received signal SIGSEGV, Segmentation fault.
0x0000000000400a7b in parse_song (song_data=0x7fffffffe7a8, attr=0x602600) at nullaby.c:43
43 *default_octave=octave_number;
(gdb)
I have no idea what I can do to fix this.
You have an int pointer
. Right.
It just mean you have a variable pointing to some memory area.
But you haven't allocated/reserved that memory area. So it can point to anything.
And it will surely point to a memory area you don't own, hence the segmentation fault.
You need to allocate memory for the pointer...
For instance:
int * default_octave = malloc( sizeof( int ) );
Or you may also use:
int default_octave_val;
int * default_octave = &default_octave_val;
Either you allocate memory to store your int (and then get a pointer to a valid memory area), or you create a pointer to an existing memory area (in the given example, a stack address).
Then you can de-reference that pointer, as it points to a valid memory area.
If it don't, you'll have a segmentation fault, or a bus error, depending on your OS.