4
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I have this function that takes a number \$N\$, and \$N\$ C strings, concatenates and returns the result:

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

char* mystrcat(int count, ...)
{
    char** p;
    char* result;
    char* ptr;

    size_t* len_array;
    va_list ap;
    int j;
    size_t total_length;

    if (count < 1)
    {
        return "";
    }

    va_start(ap, count);
    p = malloc(sizeof(char*) * count);
    len_array = calloc(count, sizeof(size_t));
    total_length = 0;

    for (j = 0; j != count; ++j)
    {
        p[j] = va_arg(ap, char*);
        total_length += (len_array[j] = strlen(p[j]));
    }

    result = malloc(sizeof(char) * (total_length + 1));
    ptr = result;

    for (j = 0; j != count; ++j)
    {
        strcpy(ptr, p[j]);
        ptr += len_array[j];
    }

    *ptr = '\0';
    va_end(ap);
    return result;
}

int main(int argc, char* argv[]) {
    puts(mystrcat(5, "Hello", ", ", "world", ", ", "friends!"));
}

As always, please tell me anything that comes to mind.

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  • 1
    \$\begingroup\$ Please note that there is nothing wrong with p = malloc(sizeof(char*) * count);. Anyone saying otherwise are just mixing in subjective opinions in their review. Some prefer the p = malloc(sizeof(*p) * count) style (I do) but that's just personal opinions. The two styles are very similar and this is not a common cause for bugs, so it need to be mentioned in a general code review. Similarly subjective: it is good style to write sizeof(char) in malloc, so that your coding style remains consistent and that the code is self-documenting. \$\endgroup\$ – Lundin Sep 28 '16 at 12:00
6
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Always return allocated string or never do it

This code here:

if (count < 1)
{
    return "";
}

is bad because you return a static string whereas the normal case returns a string allocated by malloc. If the caller then tries to free this static string, it will cause some kind of error. You can fix this by either returning an allocated empty string:

if (count < 1)
{
    return calloc(1, 1);
}

or by just returning NULL:

if (count < 1)
{
    return NULL;
}

Memory leaks

Your function never frees p or len_array. Actually, I would probably not even use temporary arrays like these. If I were to write this function, I would measure the total length on one pass, and then copy the strings on a second pass.

Rewrite

After fixing the above issues, my rewrite of your function would look like this:

char *mystrcat(int count, ...)
{
    va_list ap;
    size_t  len = 0;

    if (count < 1)
        return NULL;

    // First, measure the total length required.
    va_start(ap, count);
    for (int i=0; i < count; i++) {
        const char *s = va_arg(ap, char *);
        len += strlen(s);
    }
    va_end(ap);

    // Allocate return buffer.
    char *ret = malloc(len + 1);
    if (ret == NULL)
        return NULL;

    // Concatenate all the strings into the return buffer.
    char *dst = ret;
    va_start(ap, count);
    for (int i=0; i < count; i++) {
        const char *src = va_arg(ap, char *);

        // This loop is a strcpy.
        while (*dst++ = *src++);
        dst--;
    }
    va_end(ap);
    return ret;
}
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  • \$\begingroup\$ @LokiAstari Thanks for that. I fixed up the rewrite. \$\endgroup\$ – JS1 Sep 28 '16 at 19:37
5
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Automate consistency

    p = malloc(sizeof(char*) * count);
    len_array = calloc(count, sizeof(size_t));

If you instead say

    p = malloc(count * sizeof *p);
    len_array = calloc(count, sizeof *len_array);

You can change the types of p and len_array without having to change these calls as well.

You should also check that these functions are successful. Both return NULL on failure.

Prefer checking regions rather than points

    for (j = 0; j != count; ++j)

If you write this as

    for (j = 0; j < count; ++j)

then it is more robust in the face of future changes. For example, if you increment j an extra time in an iteration, the original might overshoot the mark. This version will always stop.

C++ iterators have to compare against single values because of the way they work. Working with a loop variable doesn't have that restriction.

Prefer memcpy over strcpy

        strcpy(ptr, p[j]);

Since you already know the length and don't need to null terminate, you can use memcpy.

        memcpy(ptr, p[j], len_array[j]);

It might even be faster, as there may be a machine instruction for copying blocks of memory efficiently. The strcpy command has to copy byte by byte, as the null could be anywhere.

Avoid memory leaks

It won't matter in a toy program like this, but you should free p and len_array before returning. Otherwise each call to mystrcat would leak memory.

    free(p);
    free(len_array);

You should also do that in the caller.

    puts(mystrcat(5, "Hello", ", ", "world", ", ", "friends!"));

Should be

    char *concatenated = mystrcat(5, "Hello", ", ", "world", ", ", "friends!");
    puts(concatenated);
    free(concatenated);

This won't matter in this program as it just ends and releases all the memory then anyway. But in a longer running program that calls mystrcat many times, this could be a problem.

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2
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  • Avoid sizeof(type). It disconnects type from the variable, and leads to double maintenance. Prefer

    p = malloc(sizeof(p[0]) * count);
    len_array = calloc(count, sizeof(len_array[0]));
    
  • len_array leaks memory.

  • len_array id strictly speaking unnecessary. Consider

    for (j = 0; i != count; ++j) {
        strcpy(ptr, p[j]);
        ptr += strlen(p[j]);
    }
    

    Of course it trades time (each strlen is evaluated twice) for space (no need for a len_array). I prefer this version because it avoids extra calloc (could be costly), and keeps the semantics of the increment clear.

  • Surely you will get plenty recommendations to test the retrun value of malloc.

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2
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Others have pointed out flaws in your code (although I'd add that sizeof(char) is 1 by definition), so I'll just offer an alternative strategy, namely to write a simple 2-string concatenation and use that if you need to join more strings:

char *mystrcat(const char * s, const char * t)
{
    size_t slen = s ? strlen(s) : 0;
    size_t tlen = t ? strlen(t) : 0;

    char *cat = malloc(slen + tlen + 1);
    if (cat) {
        memcpy(cat, s, slen);
        memcpy(cat + slen, t, tlen);
        cat[slen + tlen] = '\0';
    }
    return cat;
}

Three and four-string functions are simple extensions of this.

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