Error while writing strcat() using pointers

Question

I am trying to learn C with The C programming Language by K&R. I am trying to write a the strcat() program using pointers.

char *strcat (char *s, char *t){
    char *d;
    d = s;
    while(*s++);
    s--;
    while(*s++ = *t++);
    return d;
}

int main () {
    char *name1;
    char *name2;
    name1 = "stack" ;
    name2 = "overflow";
    printf("%s %s\n", name1, name2);
    printf("strcat out : %s", strcat(name1, name2));
    return 0;
}

But I am getting the ouput as

stack overflow
Segmentation fault

Why is it not working ? Can anybody clarify the mistake here..

You need to learn arrays and pointer, declare your `char name1[SIZE];` and `char name2[SIZE];` and replace assignment statements `name1 = "stack";` as `strcpy(name1, "stack");`, `strcpy(name2, "overflow");` — Grijesh Chauhan, Aug 31 '13 at 16:10
Read: [Difference between `char *str` and `char str[]` and how both stores in memory?](http://stackoverflow.com/questions/15177420/what-does-sizeofarray-return/15177499#15177499) — Grijesh Chauhan, Aug 31 '13 at 16:11
possible duplicate of [Implementing strcat using pointers](http://stackoverflow.com/questions/18398180/implementing-strcat-using-pointers) — Grijesh Chauhan, Aug 31 '13 at 16:12
@PeteBecker: I think for C++ the answer would be "Use `std::string` instead of `char*`, for goodness sake!" — Fred Larson, Aug 31 '13 at 22:55
@FredLarson - the answer to "why does my code crash when I write to a literal string" is not "use `std::string`. That adds overhead without getting at the root of the problem. — Pete Becker, Sep 01 '13 at 13:11

Some programmer dude · Accepted Answer · 2013-08-31T16:08:45.920

2

Because the pointers are pointers to literals, and not only are those read only the destination is not big enough to contain both the strings. For at least string1 you would want to use an array of at least big enough size to contain both strings (including the terminator). Like:

char string1[128] = "Stack";

or

char string1[128];
strcpy(string1, "Stack");

edited Aug 31 '13 at 16:08

answered Aug 31 '13 at 15:55

Some programmer dude

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so what modification should i make in this program ? – noufal Aug 31 '13 at 15:59
@noufal It's in my answer. – Some programmer dude Aug 31 '13 at 16:00
I found in the internet that we need to give declare like `char *name1 = malloc(100*sizeof(char));`. But this isn't working either .. – noufal Aug 31 '13 at 16:04
@noufal That's not correct either, since you later assign the pointer to point to the literal string, which makes you loose the original pointer returned by `malloc`. Instead you need top *copy* the literal string into the array. Please see my updated answer. – Some programmer dude Aug 31 '13 at 16:08

cpp · Answer 2 · 2013-08-31T16:03:04.180

0

In this line:

while(*s++ = *t++);

you dereference s which points to one-past-the-end of the char array "stack". This causes segmentation fault error.

edited Aug 31 '13 at 16:03

answered Aug 31 '13 at 15:55

cpp

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After `while(*s++);` is `s--;`. – johnchen902 Aug 31 '13 at 15:57
It's okay to have a pointer to one-past-the-end of an array. As @johnchen902 says, the code decrements the pointer once it's found the end. – Pete Becker Aug 31 '13 at 15:58
Thanx, I've improved my answer – cpp Aug 31 '13 at 16:06

score 0 · Answer 3 · answered Aug 31 '13 at 15:56

The program tries to append characters to the literal string "stack". Don't do that; create an array of characters that's large enough to hold the result, copy the first string into the array, and then strcat the rest of the string.

Oh, and name your function something other than strcat. That name is already taken.

score 0 · Answer 4 · answered Aug 31 '13 at 16:13

0

 char *name1;
 char *name2;
 name1 = "stack" ;
 name2 = "overflow";

==>

 char name1[20];
 char name2[10];
 memcpy(name1,"stack") ;
 memcpy(name2,"overflow");

answered Aug 31 '13 at 16:13

Lidong Guo

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Error while writing strcat() using pointers

4 Answers4

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