Segmentation fault in equating a char pointer value to some char

Question

Possible Duplicate:
Why is this C code causing a segmentation fault?

char* string = "abcd";

now when i try to change some character of this string i get segmentation fault

*string = 'p';

or

string[0] = 'p';
string[0] = 52;

Can someone please explain me the reason that why is it happening.

Thanks

Alok.Kr.

http://stackoverflow.com/questions/3108832/write-permission-for-char — sinek, Aug 03 '10 at 07:00
Also [Different string initialization yields different behavior?](http://stackoverflow.com/questions/1647273/different-string-initialization-yields-different-behavior) — caf, Aug 03 '10 at 07:19
If this is only C, then remove the C++ tag, if this is C++, try avoiding names of types in the standard library as they might be confusing (i.e. do not call your strings `string`) — David Rodríguez - dribeas, Aug 03 '10 at 07:41
"equate" is used incorrectly in the title. Variables are not "equated", they are assigned. — Olaf Seibert, Jun 17 '15 at 14:48

score 4 · Accepted Answer · answered Aug 03 '10 at 07:07

4

If you write char* string = "abcd"; the string "abcd" is stocked into the static data part of your memory and you can't modify it.

And if ou write char* string = 'p';, that's just wrong. First, you try to declare a variable with the same name (string) and, worse, you try to assign a char value to a char pointer variable. This doesn't work. Same thing : char[0] = 'p'; really means nothing to your compiler except a syntax error.

answered Aug 03 '10 at 07:07

Opera

983
1
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sorry for all the typing errors, i guess i was so sleepy that i typed all those things. – Kumar Alok Aug 03 '10 at 07:15
Ok. So as I said "string[0] = 'p';" can't work because string is static data and is read-only. – Opera Aug 03 '10 at 12:57

score 2 · Answer 2 · answered Aug 03 '10 at 06:59

2

String literals are non-modifiable in C. This has been asked and answered many times before, though it isn't too easy to search for.

answered Aug 03 '10 at 06:59

caf

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3

That's why it's a good habit to write char const * p = "hello" rather than char * p = "hello" – bradgonesurfing Aug 03 '10 at 07:24

score 1 · Answer 3 · answered Aug 03 '10 at 12:33

If you want to modify string, declare it as an array, not a pointer to a string literal.

#include <stdio.h>

int main()
{
    char string[] = "hello world";
    string[0] = 'H';
    string[6] = 'W';

    printf("%s\n", string);

    return 0;
}

Results in:

$ /tmp/hello
Hello World

Segmentation fault in equating a char pointer value to some char

3 Answers3