C tips and tricks explanation - c++

Question

Can you explain me which is the mechanism behind the next code samples (I think I know but I need second opinion):
1)--------------------------
using namespace std;
int * f(int x) {
return &x;
}
int * g(int x, int y) {
return &y;
}
int * h(int x, int y, int z) {
return &z;
}
int main() {
cout << *f(42) << endl;
int * y1 = g(43, 44);
int * y2 = g(45, 46);
cout << *y1 << ", " << *y2 << endl;
int * z1 = h(47, 48, 49);
int * z2 = h(50, 51, 52);
cout << *z1 << ", " << *z2 << endl;
return 0;
}
2)--------------------------
int *a, *b;
void f(int x) {
int i[3];
i[0] = x;
i[1] = x + 1;
i[2] = x + 2;
a = i;
}
void g(int x) {
int i[3];
i[0] = x;
i[1] = x + 1;
i[2] = x + 2;
b = i;
}
int main() {
f(1);
printf("a = {%d,%d,%d}\n", a[0], a[1], a[2]);
g(2);
printf("a = {%d,%d,%d}\n", a[0], a[1], a[2]);
}
3)--------------------------
int main() {
char * hello = "hello, world!" + 3;
char * charstring = 'h' + "ello, world!";
printf("hello=%s, charstring=%s.\n", hello, charstring);
return 0;
}
Thank you.

Answer 1

I would expect those programs to crash or do other weird things when you run them.
Example 1: The functions f, g and h are returning the memory addresses of their arguments. Note that those arguments are stored on the stack, and when the functions return, the stack is unwound and the addresses will not be valid anymore. You could get lucky and the value will still be there, but you could just as well have the program crash or return some random value that's not the value that you passed to the function.
Example 2: The functions f and g set the global variables a and b to the addresses of local variables declared in the functions. Just like in the first example, those local variables will be gone when the functions return, leaving a and b pointing to something invalid.
Example 3: This is doing weird pointer arithmetic. hello will probably point to the address of the text plus 3, so you'd probably get "lo, world!" printed for this (but it could also be different, depending on how pointer arithmetic works on your particular platform). The case with charstring is similar, only here you add 'h' (ASCII value 104 - so you're adding 104 to the pointer). This will most likely crash the program.

Answer 2

I think it's a little easier for a beginner to understand these concepts if you explain step by step what is happening in the background.
1.
cout << *f(42) << endl; // Call f with the value 42
int * f(int x) { // Push an integer, x, on the stack (x = 42)
return &x; // Return a pointer to var x
} // Pop x off the stack
// Pointer now points to a memory address that is unallocated,
// which will crash the program when it tries to use that memory,
// which it does with cout
2.
f(1); // Call f with the value 1
void f(int x) { // Push an integer, x, on the stack (x = 1)
int i[3]; // Declare an int* with space for 3 vals (local! stack!)
i[0] = x; // Define values of the array
a = i; // Set a equal to i, beginning of array
} // i is now out of scope, and since it was declared as locally,
// rather than with malloc (or new in c++), it is on the stack
// and has now been popped off, so a points to a memory address
// that the OS *should* have marked as inaccessible
3.
char * hello = "hello, world!" + 3; // hello is a char*, a pointer that
// points to the beginning of an array
// of characters. Adding 3 will increment
// the pointer three characters after the
// first character.
char * charstring = 'h' + "ello, world!"; // charstring is a char*, a pointer that
// points to the beginning of an array
// of characters. This time, it would point
// to "ello, world!". However, the addition
// of 'h' will shift the character position
// by 104 characters because that is the
// value of ascii 'h'.