Simple matter about Memset in C

Question

I am a begginner at C so please bear with me; I am aware that i can make an array statement as *c or c[];

My question is about memset:

  char str[] = "hello!";
  memset (str,'-',2);
  puts (str);

Works fine. But:

char *str = "hello!";
  memset (str,'-',2);
  puts (str);

Dont work, I know that char *str = ... is a normal array statement.

If anyone can help me with this one i Thank you!

Your Question will answer you to lot of others when comes about Pointers and Arrays. Think about something like copy by value or copy by pointer (reference how others are calling it). The same concept happens here two. `char str[] = "hello";` means that you do a copy of the Literal string `hello` inside that Array and you can manipulate that string inside there, but when you do `char *str = "hello";` you deal with a pointer now which points to that Literal string and in `C` you read/access them, but you can not write/edit them. Hope you understand the difference now. — Michi, Nov 10 '17 at 23:29

immortal · Accepted Answer · 2017-11-10T23:22:41.170

The difference here is subtle - it's where the string is stored.

char str[] = "hello!"; allocates a string on the stack, which can be updated.

char *str = "hello!"; allocates a string in the programs const data segment and sets str to point to that segment. That segment can't be manipulated and your program crashes on memory access violation.

A modern computer has a complex memory layout, and a whole set of concepts you'll need to learn at some point, such as Virtual Memory and Paging and such as Stack and Heap.

A program being loaded into memory is segmented into different sections which are loaded to different pages with different permissions. The code and global const variables are loaded to pages which have no writing permissions (only read) - .text and .rodata segments respectively - while the stack and heap are allocated on pages which can be written to but can't be executed (.data and .bss).

The literal string "hello" in the 2nd example is allocated in the const segment (.rodata), and thus can't be altered. Moreover, if you define several strings like that

char *s1 = "Hello!";
char *s2 = "Hello!";

It's very likely that s1 == s2 will be true (address comparison!)

In the 1st example an actual array is being allocated on the stack and filled with the bytes containing "hello!\0" (7 bytes). That memory CAN be manipulated since it is on the stack which is allocated on writable pages.

score 0 · Answer 2 · answered Nov 10 '17 at 23:10

0

char str[] = "hello!";
  memset (str,'-',2);
  puts (str);

it's working because str is array and its content you can modify because local arrays are stored in stack section of RAM which we can modify.

but

char *str = "hello!";
  memset (str,'-',2); //

it's not working because str is pointer and str itself stored in stack section but pointing to code(read only) section of RAM in case of linux. so you are trying to modify read only memory, thats why it will not works.

answered Nov 10 '17 at 23:10

Achal

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I want to say that 'str' a char pointer in second case and its initialized with code sction address, modifying code section will yields in access violation. Is it wrong ? – Achal Nov 10 '17 at 23:14
Ok Guys thank you very much for the quick repply, Sorry it was duplicate my bad, God bless you guys; – Mondometal Nov 10 '17 at 23:21

Simple matter about Memset in C

2 Answers2