Appending char to string (char array)

Hi there..

I have searched all over the net for this little feature, but cannot find it anywhere.. How does I append a character to a string.. It sounds very simple, but can't find help to do so anywhere..

I tried to use a string to do this this as follows, but it does not do it quite well, and looks very stupid:

char strTMP[1];
char testint2[] = "BLA BLA BLA ";
strTMP[0] = c[i]; // returns a char
strcat(testint2,strTMP);

1. you should have enough room in your buffer to append something, currently your testint2 buffer size is enough only to store the "BLA BLA BLA " string...

2. easiest way to append char - using simple assignment

Code:

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char res[10]="start ";
size_t len = strlen(res);
res[len++] = 'A'; /* we overwriting the null-character with another one */
res[len] = '\0'; /* to make the string null-terminated again */

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Quote:

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Originally Posted by vart

1. you should have enough room in your buffer to append something, currently your testint2 buffer size is enough only to store the "BLA BLA BLA " string...

2. easiest way to append char - using simple assignment

Code:

---

char res[10]="start ";
size_t len = strlen(res);
res[len++] = 'A'; /* we overwriting the null-character with another one */
res[len] = '\0'; /* to make the string null-terminated again */

---

---

I was aware of that problem thats why i used strcat, which automatically allocates extra space..
I think I have tried this but then it complains about precisely "char res[10]="start ";". Something about that a value cannot be assigned this way when the length is set.. But I'll try again then and see if it should be possible..

Make sure you have enough room to do it. Otherwise, you'll overwrite some other memory with your new '\0'.

You don't really need a function to do this per se, but if you did want one:

Code:

---

char *appendchar(char *szString, size_t strsize, char c)
{
        size_t len = strlen(szString);
        if((len+1) < strsize)
        {
                szString[len++] = c;
                szString[len] = '\0';
                return szString;
        }
        return NULL;
}

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Example usage:

Code:

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char szBuffer[BUFSIZ];

/* Note: szBuffer must have a '\0' char inside of it before this function is called. */

...

appendchar(szBuffer,sizeof(szBuffer),'x');

---

Thank you both... It seems to work..

Why did it give me an error message on the specified "char res[10]="start ";" earlier.. Well, now it seems to work..

Thanks again... :)

And I have thought about the null terminating character inside the strings, but did not think it was nessasary when I just use the only character avaliable in the chararray[1], but maybe to secure I should have used chararray[2] then..

Quote:

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Originally Posted by MacGyver

Make sure you have enough room to do it. Otherwise, you'll overwrite some other memory with your new '\0'.

You don't really need a function to do this per se, but if you did want one:

Code:

---

char *appendchar(char *szString, size_t strsize, char c)
{
        size_t len = strlen(szString);
        if((len+1) < strsize)
        {
                szString[len++] = c;
                szString[len] = '\0';
                return szString;
        }
        return NULL;
}

---

Example usage:

Code:

---

char szBuffer[BUFSIZ];

/* Note: szBuffer must have a '\0' char inside of it before this function is called. */

...

appendchar(szBuffer,sizeof(szBuffer),'x');

---

---

And the usage of this method should be stored in a char[] right..? like
char res[20];
res = appendchar(szBuffer,sizeof(szBuffer),'x');

but then I get the message.. Lvalue required in function find... what does this mean..? I'm not that strong to pointers.. :S

Not quite.

If your char array is declared this way:

Code:

---

char res[20];

---

And let's say you wanted to append the character 'C', then the usage would be like this:

Code:

---

appendchar(res,sizeof(res),'C');

---

The return value is just mainly for checking if the function was successful or not.

For example:

Code:

---

...
if(appendchar(res,sizeof(res),'C'))
{
        /* The character was appended to res*/
}
else
{
        /* The character was not appended because there was not enough room in res. */
}

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Ahh, I see.. The pointer to the method saves automatically in the sting-value given to the method (*szString). It saves in the same string because it is given as a pointer, right..?

As said I'm not that good to pointers yet.. :S

Yes, correct.

When a pointer is passed, the data it points to can be altered by the function. Since a C-string is just a pointer to the first element of an array by the time it gets to a function, any character in that string can be altered by the function and the changes will be reflected in the calling function.

Thank you very much.. Now I know a little about incomming pointers to a method as well.. :)

i have some questions about that function:
when i apply to this code:

Code:

---

int main()
{
  clrscr();
  Stack *stack=new Stack(100);
  char x[100];
  printf("Input Equation: ");
  fgets(x);
  char dest[100];

  for(int i=0;i<strlen(x);i++)
  {

        if(isdigit(x[i]))
        {
                appendchar(dest, sizeof(dest), x[i]);

        }
        else if((x[i]=='+')||(x[i]=='-')||(x[i]=='*')||(x[i]=='/'))
        {
                stack->push(x[i]);
        }
        else if(x[i]==')')
        {
                appendchar(dest, sizeof(dest), stack->pop());
        }


  }
  printf("&#37;s", dest);
  printf("\n\nPress any key to exit");
    return 0;

}

char *appendchar(char *s, size_t strsize, char c)
{
        size_t len=strlen(s);
        if((len+1) < strsize)
        {
                s[len++]=c;
                s[len]='\0';
                return s;
        }
        return NULL;
}

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what will be the result for this code?

Quote:

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Originally Posted by blacksnake

what will be the result for this code?

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Compilation errors, obviously. This isn't C

Quote:

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Originally Posted by sniper83

I was aware of that problem thats why i used strcat, which automatically allocates extra space..

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The strcat function does not allocate any extra space.

Quote:

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Originally Posted by blacksnake

what will be the result for this code?

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You were told to correct that code before. Don't use old code.