(C) How to use strncat() with **argv?

Question

int main(int argc, char * argv[]){

   char file_extension[10];

   strncat(file_extension, argv[2][5], 6);

When I do this, I get "warning: passing arg 2 of 'strncat' makes pointer from integer without a cast'. Does does anyone know how to fix this?

Also, what are you trying to achieve? What you currently have has at least two UBs inside and it also makes absolutely no sense. — , Feb 08 '13 at 19:06
I'm trying to extract the file extension from an argument. I left out most of the other code to make this simple. What is a "UB"? — user1472747, Feb 08 '13 at 19:08
@userXXX For that, why don't you use `char *extension = 1 + strchr(argv[1], '.');`? — , Feb 08 '13 at 19:09

score 3 · Answer 1 · answered Feb 08 '13 at 19:08

3

argv[2] is a char *, so that would make argv[2][5] a char

strncat accepts char * parameters only!

answered Feb 08 '13 at 19:08

Forhad Ahmed

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thank you! Then, do you know how to make a new string, say from argv[2][3] to argv[2][7]? – user1472747 Feb 08 '13 at 19:10
1

@user1472747 `memcpy()`. And don't forget the terminating `NUL`. – Feb 08 '13 at 19:11
If you want to do strncat of file_extension to a substring of argv[2], you will do this: – Forhad Ahmed Feb 08 '13 at 19:19

score 2 · Answer 2 · answered Feb 08 '13 at 19:08

2

argv[2][5] is a char only.

Try with just argv[2].

answered Feb 08 '13 at 19:08

Adrián

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score 2 · Answer 3 · answered Feb 08 '13 at 19:13

2

So if I finally managed to comprehend correctly all your ambiguous comments: it seems that you want to copy a part of a string.

const char foo[] = "abcdefghijklm";
const size_t start = 3;
const size_t len = 5;

char the_copy[len + 1];
memcpy(the_copy, foo + start, len);
the_copy[len] = 0;

answered Feb 08 '13 at 19:13

@user1472747 You're welcome. I'm not "eager". See, I was a beginner too, and it was not very long ago. Yet I didn't have to ask questions like this. Show some creativity! The documentation of the standard library functions is also helpful. And if you listen to one more advice: don't want to do everything the first day you start programming! It will take several months, years, etc. till you have the necessary experience for something complex. – Feb 08 '13 at 19:26

score 1 · Accepted Answer · edited May 23 '17 at 10:32

You have several problems:

argv[2][5] is a single character: the 6th character of the 3rd input argument, but the second argument of strncat takes a pointer to a character (i.e. a C string), not a single character
file_extension is not null-terminated to begin with (it's uninitialized), so it's Undefined Behavior to call strncat on it
The last argument to strncat is the maximum number of characters of the source string to concatenate, not the size of the output buffer -- it does not protect you from buffer overflows.
If you actually meant to write strncpy instead of strncat, then you also need to be aware that strncpy does not necessarily null-terminate the output

There are are easy fixes to these problems, but the best fix for you depends on exactly what you're going to do: is the file extension read-only? Are you constructing a new filename? What's going to happen with it?

I strongly recommend you get a good book about C strings and C programming instead of trying to blindly write string code without having a solid understanding of what it's doing, especially with the high risk of buffer overflows and memory corruption when the code is not written correctly.

You are confusing `strncpy` with `strncat` in your second and third points. — interjay, Feb 08 '13 at 19:12
Editing it to say `strncpy` doesn't help since the question uses `strncat`. — interjay, Feb 08 '13 at 19:13
@interjay: Whoops sorry, I was a little confused there, it should be fixed now — Adam Rosenfield, Feb 08 '13 at 19:17

(C) How to use strncat() with **argv?

4 Answers4