string concatenation using pointer

[_AnsHUMAN_]

Divyang Mithaiwala wrote:

Because if you have taken temp as array (char temp[XX]) then you might get error

:doh: Why so?

Divyang Mithaiwala wrote:

But a code that we are sawing has no that kind of error.

Did you trust your computer to check if there was a runtime error? :)

You need to google first, if you have "It's urgent please" mentioned in your question. ;-)_AnShUmAn_

[hrishiS]

sorry, But I really didnot get that

----------------------------- I am a beginner

[Divyang Mithaiwala]

When I was posted I haven't seen your clarification that you are facing run time error (due to no refresh page). So, If you have taken variable as char temp[10]; and try to execute *temp++ = *a++; it will give you compile time error.

---

Do not trust a computer... Always check what computer is doing regards, Divyang Mithaiwala Software Engineer

[CPallini]

Well you may allocate a predefined amount of memory (on statistical grounds) and then reallocate (for instance, in C language, you may use malloc and realloc) a bigger size if needed. BTW: you know, there are C runtime library functions for the purpose of concatenating strings. :)

If the Lord God Almighty had consulted me before embarking upon the Creation, I would have recommended something simpler. -- Alfonso the Wise, 13th Century King of Castile.
This is going on my arrogant assumptions. You may have a superb reason why I'm completely wrong. -- Iain Clarke
[My articles]

[hrishiS]

yes But I wanted to implement it with my own code....So that I learn pointer concept too..

----------------------------- I am a beginner

[hrishiS]

I used the following code.....No error but not giving any results char *a,*b; char *temp; temp=new char[5]; a="aa"; b="bb"; while(*a!='\0') { *temp++ =*a++ ; } while(*b!='\0') { *temp++= *b++ ; } *temp = '\0'; printf("%s",temp);

----------------------------- I am a beginner

[CPallini]

That's almost the solution... :rolleyes:, try:

char *a,*b;
char *temp, *dest;
dest=new char[5];
temp = dest;
a="aa";
b="bb";
while(*a!='\0')
{
*temp++ =*a++ ;
}
while(*b!='\0')
{
*temp++= *b++ ;
}
*temp = '\0';
printf("%s",dest);

of course you need now to generalize the code (ad remember to free dest when you no longer need it) :)

If the Lord God Almighty had consulted me before embarking upon the Creation, I would have recommended something simpler. -- Alfonso the Wise, 13th Century King of Castile.
This is going on my arrogant assumptions. You may have a superb reason why I'm completely wrong. -- Iain Clarke
[My articles]

[Stuart Dootson]

himangshuS wrote:

I wont be knowing the no of characters after contcatination(ie the size of the temp) in advance.

Yes you can - you need to allocate strlen(A) + strlen(B) + 1 characters, where strlen is defined as follows:

int strlen(const char* s)
{
int count;
while (*(s++))
++count;
}

Java, Basic, who cares - it's all a bunch of tree-hugging hippy cr*p

[Fatbuddha 1]

I haven't tested that nor tried something similar, but I think that the hole think wont give you the right result anyway. By incrementing the char * temp it jumps to the next position and you will lose the previous position, don't you? So if you really want to do something like that you should have a pointer to the first position of temp (I guess). So why not simply allocate mem for temp and then do a memcpy with a and b? Given memcpy temp and for b &temp[2] for example. I hope this will do the trick, but I am just guessing. Cheers

You have the thought that modern physics just relay on assumptions, that somehow depends on a smile of a cat, which isn’t there.( Albert Einstein)

[hrishiS]

Oh thanks a lot...its working fine... I have to increase my knowledge towards pointer

----------------------------- I am a beginner