What is happening behind these char array declerations?

Question

Here is the whole code:

#include <stdio.h>  //printf
#include <stdlib.h> //malloc
#include <string.h> //memcpy

void main(void)
{
    char* charMem = (char*) malloc(5 * sizeof(char));
    memcpy(charMem, "Hello", 5);

    char charArr[] = "Hello";

    printf(" charMem = %p\n", charMem);
    printf("&charMem = %p\n\n", &charMem);

    printf(" charArr= %p\n", charArr);
    printf("&charArr= %p\n\n", &charArr);

    charMem[2] = 'A';
    charArr[2] = 'A';
}

And output is:

 charMem = 00F8ABB0 //diff
&charMem = 00CFFBB0 //diff

 charArr= 00CFFBA0 //same
&charArr= 00CFFBA0 //same

What is happening behind the compiler that leads to these results?

1) How can the address and the value of charArr can be same? 2) Why the error occurres with a valid statement that compiles without any warning? — Semih SÜZEN, Apr 02 '19 at 13:28
Also a duplicate of https://stackoverflow.com/questions/164194/why-do-i-get-a-segmentation-fault-when-writing-to-a-string-initialized-with-cha — Andrew Henle, Apr 02 '19 at 13:35
@AndrewHenle My 2. question was a duplicate so i deleted it. — Semih SÜZEN, Apr 02 '19 at 13:40
Note that after `memcpy()` your `charMem` does not point to a string because `'/0'` is not accounted for ... and that your `charArr` points to the initial char of a /*readonly*/ 6-byte array. — pmg, Apr 02 '19 at 13:44

pmg · Answer 1 · 2019-04-02T13:38:36.993

2

char* charPtr = "Hello";

"Hello" is a readonly array of char. Modifying its contents is undefined behaviour.

Some programmers like to help themselves with

const char *charPtr = "Hello";

so that the compiler checks for attempted changes to the readonly array contents.

charPtr[2] = 'A'; // attempt to change a readonly/const element

edited Apr 02 '19 at 13:38

answered Apr 02 '19 at 13:31

pmg

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Thanks! That is a satisfying answer to the 2. question. – Semih SÜZEN Apr 02 '19 at 13:35

alinsoar · Accepted Answer · 2019-04-02T13:44:05.377

You declare charPtr as a pointer to char and initialize it to point to a string "Hello".

The string "Hello" is kept in a read-only region of memory, so when you try to modify the 3rd location of the string "Hello" it crashes.

Correct it to declare it so

char charPtr[] = "Hello";

Declaring it like that you will be able to mutate the array charPtr[].

Looking in the annex J.2 Undefined behavior wee see that it is undefined when

— The program attempts to modify a string literal (6.4.5).

What is happening behind these char array declerations?

2 Answers2