char pointer doesn't increment?

Question

after ptr++ pointer does not increment

  1 #include<iostream>
  2 
  3 int main() {
  4 
  5         char *ptr;
  6         char ch = 'A';
  7         ptr = &ch;
  8 
  9         std::cout << "pointer :" << &ptr << "\n";
 10         ptr++;
 11         std::cout << "pointer after ++ :" << &ptr << "\n";
 12         return 0;
 13 }

ikar$ g++ pointer_arth.cpp 
ikar$ ./a.out 
pointer :0x7ffeed9f19a0
pointer after ++ :0x7ffeed9f19a0
ikar$

@Raymond I've used pointers in C and C++ for 25+ years. I think I know how they work. Thank you. — Jesper Juhl, Nov 30 '19 at 15:14
@JesperJuhl As the answer original had like the question, https://onlinegdb.com/SyXktWxTH As you will see using just `ptr` will not work. Now he has changed his answer to the correct response. The original answer had no code and said just change to ptr. — ABC, Nov 30 '19 at 15:21
@Raymond, neither will `*ptr` because then you'll potentially be dereferencing memory that doesn't belong to your program — ForceBru, Nov 30 '19 at 15:23
@ForceBru `int *a; int b = 0; a = &b;` You said just output `a` with no pointer, that would point to the memory address. But somehow you all think `*a` is wrong. Your original answer only told him to remove the reference, your first answer was wrong until it was changed. — ABC, Nov 30 '19 at 23:43
@Raymond, yes, the first version of my answer was just as wrong as dereferencing the pointer. Indeed, your comment about testing before answering helped me understand and fix the issue with my answer. Thank you — ForceBru, Dec 01 '19 at 00:02

ForceBru · Accepted Answer · 2019-12-01T00:08:27.780

3

You're incrementing the pointer, but outputting the address of the variable that holds the pointer itself (&ptr). You should output just ptr (and format it accordingly - see edit below).

Example:

#include <iostream>

int main() {
 char data;
 char *ptr = &data;

 std::cout << "pointer:" << (unsigned long long)ptr << std::endl;
 ptr++;
 std::cout << "pointer incremented: " << (unsigned long long)ptr << std::endl;
}

Output:

pointer:140732831185615
pointer incremented: 140732831185616

Yes, printing just ptr will output garbage, so I converted the pointer to an integer (since pointers are memory addresses anyway).

As suggested in the comments, you can cast the pointer to void * when printing, which gives nicer formatting:

pointer:0x7ffee5467acf
pointer incremented: 0x7ffee5467ad0

Note how 0x7ffee5467acf == 140732745022159 != 140732831185615 - you'll get different outputs on each run because the kernel will load the executable into different places in memory.

EDIT: yes, the first version of this answer, about simply outputting ptr with std::cout << ptr, was incorrect, because the << operator is overloaded in such a way that it treats pointers to char as C-strings. Thus, that version would access potentially invalid memory and output garbage.

But the concept remains the same. Pointers to int, for example, don't have this "problem" and are printed as hexadecimal numbers, even without casting them to void *: Try it online!. The output shows that pointers are still incremented correctly by sizeof(int), which equals 4 on that machine.

edited Dec 01 '19 at 00:08

answered Nov 30 '19 at 15:00

ForceBru

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2

Weird choice of formatting (why not cast to `void*` instead? then you get nice hex output _and_ you don't have to guess whether your integer type is large enough) but the rest of the answer is sound – Lightness Races in Orbit Nov 30 '19 at 15:13
Does this this code have undefined behavior simply because `ptr` doesn't point to any valid address after `ptr++`? – Aykhan Hagverdili Nov 30 '19 at 15:14
@Ay One-past-the-end should be okay. I realise it's not an array strictly. Does that matter? Hmm. – Lightness Races in Orbit Nov 30 '19 at 15:16
@LightnessRaceswithMonica, yeah, casting to `void*` should be better, I haven't thought about that. I just happen to know that `unsigned long long` can fit a pointer on my machine, so I used that – ForceBru Nov 30 '19 at 15:17
@Ay You make a good point though - we should mention in our answers that this sort of "treat pointers like numbers" arithmetic is not as simple as it looks and shouldn't be done in general – Lightness Races in Orbit Nov 30 '19 at 15:18
@Raymond There was no substantive change to this answer, least of all one that would affect the code's well-definedness. – Lightness Races in Orbit Nov 30 '19 at 15:20
@LightnessRaceswithMonica I know that the standard allows one-past-end for arrays, but not sure if is valid in this case. – Aykhan Hagverdili Nov 30 '19 at 15:20
@Ayxan http://eel.is/c++draft/conv.lval#3.3 http://eel.is/c++draft/basic.stc.dynamic.safety#2.3 http://eel.is/c++draft/expr.add#4.2 Think you're right. – Lightness Races in Orbit Nov 30 '19 at 15:23

score 1 · Answer 2 · answered Nov 30 '19 at 16:31

1

Pointer is incremented successfully in your code. You print the address of location which hold the pointer variable. Actually, it is garbage after character -'A' if print 'ptr', you can understand and pointing to such un-handled memory location is not good.

answered Nov 30 '19 at 16:31

Build Succeeded

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Thanks for this, that was all I was trying to say. Nice write up. – ABC Nov 30 '19 at 22:34

char pointer doesn't increment?

2 Answers2