So just experimenting with pointers in C.
void inc(int *p){
++(*p);
}
int main(){
int x = 0;
int *p;
*p = x;
inc(p);
printf("x = %i",x);
}
Why is this printing "x = 0" instead of "x = 1"?
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3 Answers
Here's your error:
*p = x;
You're dereferencing p, which is unassigned, and giving it the current value of x. So x isn't changed because you didn't pass a pointer to x to your function, and dereferencing an uninitialized pointer invokes undefined behavior.
You instead want to assign the address of x to p:
p = &x;
Alternately, you can remove p entirely and just pass the address of x to inc:
inc(&x);
Comments
Because you don't set p to the address of x
Use
p = &x;
instead of
*p = x;
With *p = x you cause undefined behaviour because p has indeterminate value and points *somewhere*. But you don't know where it points and with *p = x you write the value of x to that memory location.
Comments
You need to assign the address of x to p. As mentioned in the other answers you might just want to pass in inc(&x);. No need to declare a variable and waste it like that.
