trying to concatenate: help!

I am trying to construct a string message. It should start with a BEGIN-tag and end with en END-tag. I am trying to concatenate the BEGIN-tag (declared as a #define) with a string and then with the END-tag.

Code:

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

#define BEGIN "@@"
#define END "CCC"

int main(void) {

        char *beg;
        char *end;
        int i = 0;

        beg = malloc(sizeof(BEGIN));
        end = malloc(sizeof(END));
        beg = strcpy(beg,BEGIN);
        end = strcpy(end,END);
        printf("----------------Begin of Round %i------------------\n",++i);
        printf("beg pointer before concat: %s\n", beg);
        printf("end pointer before concat: %s\n", end);
        printf("BEGIN pointer before concat: %s\n", BEGIN);
        printf("END pointer before concat: %s\n", END);
        beg = strcat(beg, "Some beautiful message");
//        beg = strcat(beg, end);
        printf("beg pointer after concat: %s\n", beg);
        printf("end pointer after concat: %s\n", end);
        printf("BEGIN pointer after concat: %s\n", BEGIN);
        printf("END pointer after concat: %s\n", END);
        printf("----------------End of Round %i------------------\n\n",i);
               
        return 0;       
}

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The output that I get is this:

Code:

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----------------Begin of Round 1------------------
beg pointer before concat: @@
end pointer before concat: CCC
BEGIN pointer before concat: @@
END pointer before concat: CCC
beg pointer after concat: @@Some beautiful message
end pointer after concat:  message
BEGIN pointer after concat: @@
END pointer after concat: CCC
----------------End of Round 1------------------

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(1) The main thing that I don't understand is the value of the end pointer after the concat call. Why is it changed to what it is.
(2) Furthermore I do not really understand why I have to use the malloc statements in this case (without them the program crashed)
(3) If you look at the code, there is one statement commented-out. If I run it, it crashes, why?

Thanks for your efforts.

I'm unclear on what you're trying to do here, but you have allocated a very small amount of memory, to try and hold the message string.

MORE bytes, please! ;)



print out the sizeof your variables, and it will become much clearer to you.

You are allocating a fixed amount of memory for "beg", but then adding more characters to it without reallocating more memory to hold them.

You also aren't "free()"ing your allocated memory at the end of the program.

You might also have better luck with:

Code:

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  beg = malloc(strlen(BEGIN)+1);  // plus one for null character
  end = malloc(strlen(END)+1);    // plus one for null character

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Quote:

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Originally Posted by django

I am trying to construct a string message. It should start with a BEGIN-tag and end with en END-tag. I am trying to concatenate the BEGIN-tag (declared as a #define) with a string and then with the END-tag.

---

Like this....

Code:

---

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

#define BEGIN "@@"
#define END "CCC"

int main (void)
  {  char *result = calloc(100,sizeof(char));  // create some string space

    strcat(result, BEGIN);                      // result holds @@
    strcat(result, "this is a message");    //  result holds @@this is a message
    strcat(result, END);                        // result holds @@this is a messageCCC

    printf("%s\n\n", result);
    free(result);
    return 0; }

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Alll so nice and simple...

Tater thanks, way more simple and elegant. Still you would help my understanding if you could answer (1) from the start of this post. Then I'll be ready to abandon this wreck...

I tested a bit more. The end pointer seems to be overwritten because I allocate a string to the beg pointer which is way longer than the space allocated. If I augment the malloc of beg, the end pointer gives the value I would expect. Does this mean that if in C you load more into a pointer segment than you allocated, it can be screwing up other pointers and so your whole program?

You don't seem to understand concatenating.

You have two words: "Bob" and "Bobberson". You want them to be one word. But the space for the first word just holds "Bob", plus the nul character at the end. You have to make more space so that you can stick them together.

length of "Bob" + one space between them + length of "Bobberson" + 1 nul character at the end = size of new space.

Make that space. Copy "Bob" into it. Stick a space on the end. Now stick "Bobberson" onto the end. There, you have successfully concatenated "Bob" " " "Bobberson" into the same chunk of memory.

Or, if the space for "Bob" was big enough for both words in the first place, you can skip the step of making new space. Just copy space + "Bobberson" onto the end. In fact, that is what strcat(s,append) assumes: s has enough space for append + a nul character.

If you have made new space for it, and have copied both words there, you still have your old words "Bob" and "Bobberson" floating around, unless you used realloc on "Bob" (assuming that you actually could - ie: it's not a string literal). You are just making new space, or resizing the current space, and then copying what was in some old space into that new space.

It has nothing to do with how many pointers you make. You don't concatenate by doing:

Code:

---

ptr1 = "Bob"
ptr2 = "Bobberson"
ptr1 = ptr1 + ptr2;

---

If that is what you are trying to say in the previous post.


Quzah.

No, that's not what I am saying. I simply would like an answer to question (1) of my original post.

Quote:

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Originally Posted by django

No, that's not what I am saying. I simply would like an answer to question (1) of my original post.

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It doesn't matter what it is, because you are invoking undefined behavior:

Code:

---

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

#define BEGIN "@@"
#define END "CCC"

int main(void) {

        char *beg;
        char *end;
        int i = 0;

        beg = malloc(sizeof(BEGIN)); /* 3 characters worth of space */
        end = malloc(sizeof(END));  /* 3 characters worth of space */
        beg = strcpy(beg,BEGIN);
        end = strcpy(end,END);
        printf("----------------Begin of Round %i------------------\n",++i);
        printf("beg pointer before concat: %s\n", beg);
        printf("end pointer before concat: %s\n", end);
        printf("BEGIN pointer before concat: %s\n", BEGIN); /* not a pointer, a string literal */
        printf("END pointer before concat: %s\n", END); /* not a pointer, a string literal */
        beg = strcat(beg, "Some beautiful message"); /* you don't have space to do that */
        /* nothing here matters, because you just trashed
            whatever memory was beyond the space
            allocated for 'beg' */

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Running off the end of your allocated space is undefined behavior. The fact that your program actually runs at this point is pure luck. Any values you have anywhere are irrelevant at this point.


Quzah.

Ok, thanks. I am leaving this thread, my initial question was too ill-formed anyway. Thanks for the comments.

Quote:

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Originally Posted by django

I tested a bit more. The end pointer seems to be overwritten because I allocate a string to the beg pointer which is way longer than the space allocated. If I augment the malloc of beg, the end pointer gives the value I would expect. Does this mean that if in C you load more into a pointer segment than you allocated, it can be screwing up other pointers and so your whole program?

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That's correct, you only allocate sizeof(pointer) for your string space. Most likely, being consecutive calls, the memory exists tail to nose and when you write your message you are overwriting the memory for your end string.

In any case... no matter what's happening there... THAT is not the way to do it.