Weird bug in pointers

Question

I was trying to access single bytes of an int value via the code below. My problem is that whenever I try to remove long int i=0; from the code, it gives me a segmentation fault. Is there any reason this happens? I am not using I anywhere in the code.

// Online C++ compiler to run C++ program online
#include <iostream>

int main() {
    // Write C++ code here
    
    unsigned int* a; 
    
    unsigned char* b1;
    unsigned char* b2; 
    unsigned char* b3; 
    unsigned char* b4;
    
    *a= 4294967295; //set to max val (4 bytes)
    //*************************
    long int i=0;//Q. Why long int/long long int? 
    //*************************
    
    b1 = (unsigned char*)(a);
    b2 = b1+(long long int)1;
    b3 = b1+(long long int)2;
    b4 = b1+(long long int)3;
    
    std::cout <<*a<<" "<<(int)*b1<<" "<<(int)*b2<<" "<<(int)*b3<<" "<<(int)*b4<<std::endl;
    
    
    return 0;
}

The pointer `a` is never initialized to point anywhere. So when you dereference it you will have *undefined behavior*. Simple solution: Make `a` a value instead of a pointer, as in `unsigned int a = 4294967295;` — Some programmer dude, May 06 '22 at 08:09

score 3 · Accepted Answer · answered May 06 '22 at 08:13

3

This exhibits undefined behavior:

unsigned int* a; 
*a= 4294967295; //set to max val (4 bytes)

The pointer variable a is never initialized to anything, so it points to a random memory address. Writing anything to that random garbage address (typically) causes a segmentation fault. It's just coincidence that adding another variable changes the behavior (due to a change in memory layout of the program).

answered May 06 '22 at 08:13

PMF

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perhaps worth mentioning that `*b2` and others are UB as well. Pointer arithmetics is only defined for pointers into an array. – 463035818_is_not_an_ai May 06 '22 at 08:15
1

@463035818_is_not_a_number Even for `unsigned char*`? How can a user re-implement `memcpy` then? – Quimby May 06 '22 at 08:21
1

@463035818_is_not_a_number I don't quite understand. `unsigned char*` is (also) an array type. – PMF May 06 '22 at 08:22
@PMF oh sorry, I was wrong. Once `a` is fixed the `bx` are allowed to alias it because they are char pointers – 463035818_is_not_an_ai May 06 '22 at 08:28

score 0 · Answer 2 · answered May 06 '22 at 08:53

The posted program has Undefined Behavior⁽¹⁾:

unsigned int* a;  // Uninitialized local variable, it has an indeterminated value. 
*a = 4294967295;  // Where is it assigned?

It's unfortunate that it happens to "work" in your environment, you should enable more warnings.

Since C++20, you can use std::bit_cast to reinterpret an object representation:

#include <array>
#include <bit>
#include <iostream>

int main()
{
  auto a{ 4294967295LL };
  using repr_t = std::array<std::byte, sizeof(a)>;
  
  auto b{ std::bit_cast<repr_t>(a) };
  
  // The following outputs:      4294967295   255 255 255 255 0 0 0 0
  std::cout << a << "  ";
  for (auto i : b)
    std::cout << ' ' << static_cast<int>(i);
  std::cout << '\n';    
}

^{(1) See e.g.:
Undefined, unspecified and implementation-defined behavior
Where exactly does C++ standard say dereferencing an uninitialized pointer is undefined behavior?}

Weird bug in pointers

2 Answers2