Editing Segmentation fault (section)

=== Writing to read-only memory ===
Writing to read-only memory raises a segmentation fault. At the level of code errors, this occurs when the program writes to part of its own [[code segment]] or the read-only portion of the [[data segment]], as these are loaded by the OS into read-only memory.

Here is an example of [[ANSI C]] code that will generally cause a segmentation fault on platforms with memory protection. It attempts to modify a [[string literal]], which is undefined behavior according to the ANSI C standard. Most [[compiler]]s will not catch this at compile time, and instead compile this to executable code that will crash:

<syntaxhighlight lang=c>
int main(void)
{
    char *s = "hello world";
    *s = 'H';
}
</syntaxhighlight>

When the program containing this code is compiled, the string "hello world" is placed in the [[rodata]] section of the program [[executable file]]: the read-only section of the [[data segment]]. When loaded, the operating system places it with other strings and [[constant (programming)|constant]] data in a read-only segment of memory. When executed, a variable, ''s'', is set to point to the string's location, and an attempt is made to write an ''H'' character through the variable into the memory, causing a segmentation fault. Compiling such a program with a compiler that does not check for the assignment of read-only locations at compile time, and running it on a Unix-like operating system produces the following [[runtime error]]:

<syntaxhighlight lang="console">
$ gcc segfault.c -g -o segfault
$ ./segfault
Segmentation fault
</syntaxhighlight>

[[Backtrace]] of the core file from [[GDB]]:

<syntaxhighlight lang=c>
Program received signal SIGSEGV, Segmentation fault.
0x1c0005c2 in main () at segfault.c:6
6               *s = 'H';
</syntaxhighlight>

This code can be corrected by using an array instead of a character pointer, as this allocates memory on stack and initializes it to the value of the string literal:
<syntaxhighlight lang=c>
char s[] = "hello world";
s[0] = 'H';  // equivalently, *s = 'H';
</syntaxhighlight>
Even though string literals should not be modified (this has undefined behavior in the C standard), in C they are of <code>static char []</code> type,<ref>{{cite book|title=ISO/IEC 9899:1990 - Programming languages -- C|section=6.1.4 String literals}}</ref><ref>{{cite book|title=ISO/IEC 9899:1999 - Programming languages -- C|section=6.4.5 String literals}}</ref><ref>{{cite book|title=ISO/IEC 9899:2011 - Programming languages -- C|section=6.4.5 String literals|url=http://www.iso-9899.info/n1570.html#6.4.5p6}}</ref> so there is no implicit conversion in the original code (which points a <code>char *</code> at that array), while in C++ they are of <code>static const char []</code> type, and thus there is an implicit conversion, so compilers will generally catch this particular error.