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Editorial Team
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Editorial Team
Asked: June 17, 20262026-06-17T19:09:17+00:00 2026-06-17T19:09:17+00:00

I read this on wikipedia int main(void) { char *s = hello world; *s

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I read this on wikipedia

    int main(void)
 {

    char *s = "hello world";
    *s = 'H';

 }

When the program containing this code is compiled, the string “hello world” is placed in the section of the program executable file marked as read-only; when loaded, the operating system places it with other strings and 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**

i don’t know why the string is placed in read only segment.please someone could explain this.

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  1. Editorial Team
    Editorial Team
    2026-06-17T19:09:18+00:00Added an answer on June 17, 2026 at 7:09 pm

    String literals are stored in read-only memory, that’s just how it works. Your code uses a pointer initialized to point at the memory where a string literal is stored, and thus you can’t validly modify that memory.

    To get a string in modifiable memory, do this:

    char s[] = "hello world";
    

    then you’re fine, since now you’re just using the constant string to initialize a non-constant array.

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