Segmentation Fault With Char Array and Pointer in C on Linux

Question

So I have the following program:

int main(){
  char* one = "computer";
  char two[] = "another";
  two[1]='b';
  one[1]='b';
  return 0;
}

It segfaults on the line "one[1]='b'" which makes sense because the memory that the pointer "one" points to must be in read only memory. However, the question is why doesn't the line "two[1]='b'" segfault? Looking at the assembly output from gcc:

.file   "one.c"
        .section        .rodata
.LC0:
        .string "computer"
.LC1:
        .string "another"
        .text
.globl main
        .type   main, @function
main:

We see that both strings are in the rodata section so they are readonly. So then how come the line "two[1]='b' does not segfault?

Answer 1

one points directly to the string located in a read-only page. On the other hand, two is an array allocated on the stack and is initialized with some constant data. At run time, the string in the read only section of the executable will be copied to the stack. What you are modifying is the copy of that string on the stack, not the read-only memory page.

At a higher level perspective, from the language point of view, "abcd" is an expression of type const char* and not char*. Thus, modifying the value pointed by such an expression results in undefined behavior. The statement char* one = "something"; merely stores the pointer to the string in a variable (unsafely, since it's casting away const modifier). The char two[] = "something"; is totally different. It's actually declaring an array and initializing it, much like int a[] = {1,2,3};. The string in quotes here is the initialization expression.