I'm taking CS 107 Stanford course, and in the lecture we have this simple stack struct
// stack.h
typedef struct {
int *elems;
int logicallen;
int alloclen;
} stack;
he mentioned by doing
stack *s;
it should automatically reserve memory space for the struct.
However, I got a segmentation fault and I have to manually allocate memory by doing this:
stack *s = malloc(3*sizeof(int));
When I try to print s, it shows a 0
// this will cause segmentation fault
int main(){
stack *s;
printf("%d\n", s);
StackNew(s);
}
// this is fine
int main(){
stack *s = malloc(sizeof(stack));
printf("%d\n", s);
StackNew(s);
}
So what exactly does stack *s; do?
he mentioned by doing
stack *s;it should automatically reserve memory space for the struct.
No, it should never do that.
So what exactly does
stack *s;do?
What it does however is declare a pointer to a stack. Or, in other words, reserve memory space for a pointer to stack. That pointer is not initialized, meaning it points to some random (*) memory location. And attempting to dereference it will result in undefined behaviour, which means that anything can happen, including your code seemingly working. This explains your statement:
When I try to print s, it shows a 0
It's good practice to almost always initialize your pointers to NULL prior using them. This way, you will avoid a hole category of problems.
And also, don't forget to free your allocated resources after you no longer need them.
(*) Generally, it points to wherever it was pointing to the last time it was used.