why does this not work?

[Matt Kowalczyk]

How come this function doesn't work?

void shift(char** list, int index) {
char buff = (*list)[0];
memmove((*list)+1, *list, index);
(*list)[0] = buff;
return;
}

I call it like so:

char* t = "abcde";

shift(&t, 1);

I expect t = "bacde"

or

shift(&t, 0) would not effect t at all.

Anyway, I made it work by creating another character array inside the function.
I then did a memcpy to copy list to this temp buffer and the memmove function
worked correctly when I passed in the temp buffer local to the function. I
would like to understand why the above code will not work. Am I doing something
wrong?

Thanks,
Matt

 

[Gordon Burditt]

How come this function doesn't work?

If you attempt to write on a quoted string literal, you invoke
the wrath of undefined behavior.

What *does* it do on your machine? Smegmentation fault?

void shift(char** list, int index) {
char buff = (*list)[0];
memmove((*list)+1, *list, index);
(*list)[0] = buff;
return;
}

I call it like so:

char* t = "abcde";

shift(&t, 1);

I expect t = "bacde"

or

shift(&t, 0) would not effect t at all.

Anyway, I made it work by creating another character array inside the function.
I then did a memcpy to copy list to this temp buffer and the memmove function
worked correctly when I passed in the temp buffer local to the function. I
would like to understand why the above code will not work. Am I doing something
wrong?

You are attempting to write on a string literal. That's bad.

Gordon L. Burditt

 

[pete]

How come this function doesn't work?

If you attempt to write on a quoted string literal, you invoke
the wrath of undefined behavior.

What *does* it do on your machine? Smegmentation fault?

void shift(char** list, int index) {
char buff = (*list)[0];
memmove((*list)+1, *list, index);
(*list)[0] = buff;
return;
}

I call it like so:

char* t = "abcde";

shift(&t, 1);

I expect t = "bacde"

or

shift(&t, 0) would not effect t at all.

Anyway,
I made it work by creating another character array
inside the function.
I then did a memcpy to copy list to this temp buffer
and the memmove function
worked correctly when I passed in the temp buffer
local to the function.
I would like to understand why the above code will not work.
Am I doing something
wrong?

You are attempting to write on a string literal. That's bad.

Also,
the function is overly complicated
concerning levels of indirection,
and bear in mind that it is undefined
if strlen(t) is less than 2.

/* BEGIN new.c */

#include <stdio.h>
#include <string.h>

void shift(char* list, size_t index);

int main (void)
{
char t[] = "abcde";

shift(t, 1);
puts(t);
return 0;
}

void shift(char* list, size_t index)
{
char buf = list[1];

memmove(list + 1, list, index);
list[0] = buf;
}

/* END new.c */

 

[Matt Kowalczyk]

How come this function doesn't work?

If you attempt to write on a quoted string literal, you invoke
the wrath of undefined behavior.

What *does* it do on your machine? Smegmentation fault?

void shift(char** list, int index) {
char buff = (*list)[0];
memmove((*list)+1, *list, index);
(*list)[0] = buff;
return;
}

I call it like so:

char* t = "abcde";

shift(&t, 1);

I expect t = "bacde"

or

shift(&t, 0) would not effect t at all.

Anyway, I made it work by creating another character array inside the function.
I then did a memcpy to copy list to this temp buffer and the memmove function
worked correctly when I passed in the temp buffer local to the function. I
would like to understand why the above code will not work. Am I doing something
wrong?

You are attempting to write on a string literal. That's bad.

Gordon L. Burditt

It generated a segmentation fault. I thought that I would need to pass in a
poniter to the string that I was modifying for the changes to effect to the
string outside the function.

I'm going to try the code that pete posted to see if it will work. I still
don't understand why my function wasn't working. Could you explain writing on a
string literal?

Assuming pete's code works with this function signature:

void shift(char* list, size_t index)

why would it fail with my signature? I'm just passing a pointer to the string.

void shift(char** list, size_t index)?

/* BEGIN pete's code */

#include <stdio.h>
#include <string.h>

void shift(char* list, size_t index);

int main (void)
{
char t[] = "abcde";

shift(t, 1);
puts(t);
return 0;
}

void shift(char* list, size_t index)
{
char buf = list[1];

memmove(list + 1, list, index);
list[0] = buf;
}

/* END pete's code /*

 

[Default User]

If you attempt to write on a quoted string literal, you invoke
the wrath of undefined behavior.

I'm sorry, I don't know who you are talking to because don't have
proper attributions. Please try to be responsible participant in the
newsgroup.




Brian

 

[Ben Bacarisse]

Could you explain writing on a
string literal?

It is simply part of the specification of the language that you may not
modify a string literal.

char *s = "abcde";
s[1] = 'x'; /* not allowed: t points to a string literal */

char t[] = "abcde"; /* the array t is initialised with "abcde" */
t[1] = 'x'; /* OK because t is an ordinary array */