my own strcat function using C errors

Question

I'm basically coding my very own string functions in C.

I've been trying to do the strcat function using pointers and cannot seem to understand whether I should be allocating memory using malloc or leaving it up to the heap.

char *my_strcat(const char *s1, const char *s2)
{


    const unsigned char *p1 = (const unsigned char *)s1;
    const unsigned char *p2 = (const unsigned char *)s2;

    unsigned char *string;
        //string = malloc(strlen(s1) + strlen(s2) + 1);
    while (*p1 != '\0')
{
        *string = *p1;
        string++;
        p1++;

        if(*p1 == '\0')
        {
            while(*p2 != '\0')
            {
                *string = *p2;
                string++;
                p2++;
            }
        }
    }
    return (char *)string;  
}

Any tips on more efficiently performing this task or things I'm doing wrong would be great!

Cheers

EDIT

OK so I got a working solution but just wondering after I use malloc where should I free() it?

char *my_strcat(const char *s1, const char *s2)
{


    const unsigned char *p1 = (const unsigned char *)s1;
    const unsigned char *p2 = (const unsigned char *)s2;

    char *string = malloc(sizeof(char *));
    char *res = string;

    while (*p1 != '\0')
{
        *string = *p1;
        string++;
        p1++;
    }
    while (*p2 != '\0')
    {
        *string = *p2;
        string++;
        p2++;
}
    *string = '\0'; 

    return (char *)res; 
}

you'd be better off implementing strncat... strcat() is a major major source of stack overflow holes in apps. — Marc B
– Marc B, Commented Aug 29, 2012 at 3:49
How do you get a Stack Overflow from strcat()? Buffer overruns I understand, but... — mimicocotopus
– mimicocotopus, Commented Aug 29, 2012 at 3:53
string needs to point to the end of p2. PS not initializing string does not "leave it to the heap". — obataku
– obataku, Commented Aug 29, 2012 at 3:57
@MarcB: the standard strncat() function has an interface designed to trigger errors; it is appalling. — Jonathan Leffler
– Jonathan Leffler, Commented Aug 29, 2012 at 4:07
@JonathanLeffler I am appending a string to the end of a destination string. I think if you look closer I am roughly doing the right as this works with arrays... I asked for tips... — Alistair Gillespie
– Alistair Gillespie, Commented Aug 29, 2012 at 5:37

Sergey Kalinichenko · Accepted Answer · 2012-08-29 03:53:14Z

3

First, I assume that the allocation is commented out by mistake.

You need to save the pointer that you allocate, and return it. Otherwise, you're returning a pointer string, which points at the end of the concatenation result
You are not terminating the resultant string; you need to add *string = '\0'
You should move the second loop to the outside of the first loop, and drop the if condition around it: if the first loop has terminated, you know that *p1 points to \0

char *string = malloc(strlen(s1) + strlen(s2) + 1);
char *res = string;
for (; *p1 ; *string++ = *p1++);
for (; *p2 ; *string++ = *p2++);
*string = '\0';
return res;

answered Aug 29, 2012 at 3:53

Sergey Kalinichenko

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6 Comments

Jonathan Leffler Over a year ago

The standard strcat() function does no memory allocation.

Nathan Andrew Mullenax Over a year ago

Also: malloc can return NULL, so it might be a good idea to check for that. If you get a warning for the char *string = malloc... line, an explicit cast to (char*) will make it go away.

Alok Singhal Over a year ago

@Nathan stackoverflow.com/questions/953112/… stackoverflow.com/questions/605845/…

Nathan Andrew Mullenax Over a year ago

@Alok. Solid. Nice catch. To summarize the links: the warning is from not including stdlib.h. Void pointers are automatically promoted to pointer to whatever.

Sergey Kalinichenko Over a year ago

@JonathanLeffler I understand that the standard strcat does not allocate memory, but it looks like the "OP's very own" my_strcat was going to, so I went along with the OP on that.

|

mimicocotopus · Accepted Answer · 2012-08-29 03:51:20Z

2

strcat doesn't allocate any memory so if you're trying to accomplish the same thing then you don't need to use malloc.

char* strcat(char* destination,char* source) {
 int c = 0;
 int sc;

 while(destination[c] != 0) {  c++; }

 for(sc = 0;sc < strlen(source);sc++) {
  destination[sc+c] = source[sc];
 }

 destination[sc+c] = 0;

 return destination;

}

answered Aug 29, 2012 at 3:51

mimicocotopus

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Comments

Ankuraho · Accepted Answer · 2013-11-12 16:40:16Z

0

This worked for me.

char* my_strcat(char* a,char* b)
{
        int i,j;

        for(i=0;a[i];i++);
        for(j=0;b[j];j++,i++)
         a[i]=b[j];
        a[i]='\0';

}

answered Nov 12, 2013 at 16:40

Ankuraho

1

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