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From: Bart <bc@freeuk.com>
Newsgroups: comp.lang.c
Subject: Re: relearning C: why does an in-place change to a char* segfault?
Date: Fri, 2 Aug 2024 10:43:36 +0100
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On 02/08/2024 02:06, Kaz Kylheku wrote:
> On 2024-08-01, Bart <bc@freeuk.com> wrote:
>>>> It segfaults when the string is stored in a read-only part of the binary.
>>>
>>> A string literal creates an array object with static storage duration.
>>> Any attempt to modify that array object has undefined behavior.
>>
>> What's the difference between such an object, and an array like one of
>> these:
> 
> Programming languages can have objects that have the same lifetime, yet some
> of which are mutable and some of which are immutable.
> 
> If the compiler believes that the immutable objects are in fact
> not mutated, it's a bad idea to modify them behind the compiler's
> back.
> 
> There doesn't have to be any actual difference in the implementation of
> these objects, like in what area they are stored, other than the rules
> regarding their correct use, namely prohibiting modification.
> 
> The Racket language has both mutable and immutable cons cells.
> The difference is that the immutable cons cells simply lack the
> operations needed to mutate them. I'm not an expert on the Racket
> internals but I don't see a reason why they couldn't be stored in the
> same heap.
> 
>>    static char A[100];
>>    static char B[100]={1};
>>
>> Do these not also have static storage duration? Yet presumably these can
>> be legally modified.
> 
> That 1 which initializes B[0] cannot be modified.
> 

Why not? I haven't requested that those are 'const'. Further, gcc has no 
problem running this program:

     static char A[100];
     static char B[100]={1};

     printf("%d %d %d\n", A[0], B[0], 1);
     A[0]=55;
     B[0]=89;
     printf("%d %d %d\n", A[0], B[0], 1);

But it does use readonly memory for string literals.

(The point of A and B was to represent .bss and .data segments 
respectively. A's data is not part of the EXE image; B's is.

While the point of 'static' was to avoid having to specify whether A and 
B were at module scope or within a function.)

 > That 1 which initializes B[0] cannot be modified.

Or do you literally mean the value of that '1'? Then it doesn' make 
sense; here that is a copy of the literal stored in one cell of 'B'. The 
value of the cell can change, then that particular copy of '1' is lost.

Here:

     static char B[100] = {1, 1, 1, 1, 1, 1};

changing B[0] will not affect the 1s in B[1..5], and in my example 
above, that standalone '1' is not affected.