Malloc inside a function call appears to be getting freed on return?
c, malloc, pointers
Solution
You pass the pointer by value, not by reference, so whatever you do with arr inside foo will not make a difference outside the foo-function. As m_pGladiator wrote one way is to declare a reference to pointer like this (only possible in C++ btw. C does not know about references):
int main(int argc, char ** argv) {
int * arr;
foo(arr);
printf("car[3]=%d\n",arr[3]);
free (arr);
return 1;
}
void foo(int * &arr ) {
arr = (int*) malloc( sizeof(int)*25 );
arr[3] = 69;
}
Another (better imho) way is to not pass the pointer as an argument but to return a pointer:
int main(int argc, char ** argv) {
int * arr;
arr = foo();
printf("car[3]=%d\n",arr[3]);
free (arr);
return 1;
}
int * foo(void ) {
int * arr;
arr = (int*) malloc( sizeof(int)*25 );
arr[3] = 69;
return arr;
}
And you can pass a pointer to a pointer. That's the C way to pass by reference. Complicates the syntax a bit but well - that's how C is...
int main(int argc, char ** argv) {
int * arr;
foo(&arr);
printf("car[3]=%d\n",arr[3]);
free (arr);
return 1;
}
void foo(int ** arr ) {
(*arr) = (int*) malloc( sizeof(int)*25 );
(*arr)[3] = 69;
}
Problem
I think I've got it down to the most basic case: ``` int main(int argc, char ** argv) { int * arr; foo(arr); printf("car[3]=%d\n",arr[3]); free (arr); return 1; } void foo(int * arr) { arr = (int*) malloc( sizeof(int)*25 ); arr[3] = 69; } ``` The output is this: ``` > ./a.out car[3]=-1869558540 a.out(4100) malloc: *** error for object 0x8fe01037: Non-aligned pointer being freed *** set a breakpoint in malloc_error_break to debug > ``` If anyone can shed light on where my understanding is failing, it'd be greatly appreciated.