Char array and char *

If I have an array of char and a char pointer, how can I make the compatible ?

Code:

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void something(char *mstring)
{
        char list[i] = ... // "Hello World"
                mstring = list;  ?????  // Now I want mstring to return "Hello world"
}

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Best wishes, Desmond

It sounds like you want to copy from one string to another:

Code:

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void something(char *mstring)
{
    char list[] = "Hello World";
    strcpy(mstring, list);
}

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or perhaps a little safer:

Code:

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void something(char *mstring, size_t len)
{
    char list[] = "Hello World";
    strncpy(mstring, list, len);
}

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How are you calling your function? It is unclear form your example what you want the function to do.

If you want to modify a char pointer in the calling routine, then you need to pass a char **. However, in that case you must either make list a const char pointer to a literal string, or static.

If you want to copy the string created in list to a buffer passed to you function, then you can use the strcpy() function found in <string.h>. However, you must first make sure the buffer is big enough.

Quote:

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or perhaps a little safer:

Code:

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void something(char *mstring, size_t len)
{
    char list[] = "Hello World";
    strncpy(mstring, list, len);
}

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Remember that strncpy() doesn't add a terminating NULL if there are at least len characters in the string . . . in other words, this is closer to being an equivalent of the first something():

Code:

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void something(char *mstring, size_t len)
{
    char list[] = "Hello World";
    strncpy(mstring, list, len);
    mstring[len] = 0;
}

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Also note that list could be of type const char * and it would still work. It would probably use less memory, too. ;)

Quote:

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mstring[len] = 0;

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isn't it out of bounds access?

Quote:

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Originally Posted by vart

isn't it out of bounds access?

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Depends whether len is the number of characters in the string before the null terminator(like what you would expect strlen to return) or the amount of memory allocated for mstring.

Quote:

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Originally Posted by Brad0407

Depends whether len is the number of characters in the string before the null terminator(like what you would expect strlen to return) or the amount of memory allocated for mstring.

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I know it, but in most cases when passing the buffer with len, I do it as

Code:

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somefunc(buffer, sizeof buffer);

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Generally, I think of string lengths in terms of strlen(). You know . . . buffer[strlen(buffer)] . . . malloc(strlen(buffer) + 1) . . . .

On the other hand there is fgets(), of course. :) But I have a convincing argument for my point of view!

Code:

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    strncpy(mstring, list, len);
    mstring[len] = 0;

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strncpy() takes the maximum number of characters to copy -- in other words, a strlen()-like value. Thus, strncpy() could fill in mstring[0] ... mstring[len-1]. In order to avoid overwriting any of those characters, mstring[len] is the next logical index to use.

It's just an idiom -- you know:

Code:

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strncpy(a, b, n);
a[n] = 0;

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It's reasonably common . . . it's even in the example in the link I posted earlier in this post.

But it doesn't matter, it's just an example. That's why it's called something(). ;)