Does pointer qtr hold the same or new memory in every function call in my code
I am confused because qtr is a local variable. local variable is create and destroy in function. I think i qtr should hold the same in every function callCode:#include<stdio.h>#include<stdlib.h> void foo() { int *qtr = malloc (sizeof(int)); if (qtr != NULL) { *qtr = 10; } } int main() { foo(); foo(); foo(); return 0; }
qtr will receive a different address in all three calls to foo(). You are then throwing away the pointer returned from malloc(), resulting in lost memory.
If you malloc() memory in a function, you should either free the memory before exiting the function, or return the pointer to the calling function, and have the calling function deal with freeing the memory.
You should also look at a utility called valgrind, and use it for debugging memory errors.
Last edited by rstanley; 01-05-2023 at 11:00 AM.
> I think i qtr should hold the same in every function call
While you're busy thinking about the answer, also think about writing some tests to test your ideas.
Since foo is always called from the same parent, local variables will be at the same address.Code:#include<stdio.h> #include<stdlib.h> void foo() { int *qtr = malloc (sizeof(int)); printf("DEBUG: address of variable=%p, value of variable=%p\n", (void*)&qtr, (void*)qtr); if (qtr != NULL) { *qtr = 10; } } int main() { foo(); foo(); foo(); return 0; } DEBUG: address of variable=0x7fffccfc3400, value of variable=0x5599596702a0 DEBUG: address of variable=0x7fffccfc3400, value of variable=0x5599596706d0 DEBUG: address of variable=0x7fffccfc3400, value of variable=0x5599596706f0
The answer it gets back from malloc will always be different (since you never called free).
If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
If at first you don't succeed, try writing your phone number on the exam paper.
I am trying to modify value of structure via function call. I don't understand how to do it in function
I am passing address of structure to function
OutputCode:#include<stdio.h> #include<stdlib.h> struct node { int x ; char y; }; void modify( struct node **dsptr) { printf(" dsptr store at location : %p \n", &dsptr); printf(" value of dsptr = %p \n", dsptr); printf(" %p \n", *dsptr); printf(" %p \n", **dsptr); } int main(void) { struct node *sptr = malloc (sizeof(struct node)); printf(" sptr store at location : %p \n", &sptr); printf(" value of sptr = %p \n", sptr); sptr -> x = 1; sptr -> y = 'A'; printf(" value of first object of structure = %d \n", sptr -> x); printf(" value of second object of structure = %c \n", sptr -> y); modify(&sptr); return 0; }
Code:sptr store at location : 0061FF1C value of sptr = 009F1598 value of first object of structure = 1 value of second object of structure = A dsptr store at location : 0061FF00 value of dsptr = 0061FF1C 009F1598 00000001
You don't need the extra indirection.
Code:void modify( struct node *dsptr) { dsptr->x = 42; dsptr->y = 'Q'; } int main(void) { struct node *sptr = malloc (sizeof(struct node)); printf(" sptr store at location : %p \n", &sptr); printf(" value of sptr = %p \n", sptr); sptr -> x = 1; sptr -> y = 'A'; printf(" value of first object of structure = %d \n", sptr -> x); printf(" value of second object of structure = %c \n", sptr -> y); modify(sptr); return 0; }
If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
If at first you don't succeed, try writing your phone number on the exam paper.
