concat two char * variables

[Fatbuddha 1]

Yes, I was wondering too

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)

[santhosh padamatinti]

My interviewer thrown a question like: He wants to implement a function where he needs to takes two char* parameters as input at the end it should return a char* by concatenating these two char* values(without using strcat(), strcmp() functions). required function prototype is

char\* concat(const char\* c1, const char\* c2)

I am unable to solve this one. Please give me the approach Thanx in advance.....

To invent something, you need a mountain of junk in your mind. ---------------------Thomas alva edison

modified on Thursday, January 21, 2010 2:21 AM

[CPallini]

Why wondering? It is concise and elegant, functionally equivalent to strcat, here the complete function

char * concat( char * c1, const char * c2)
{
char *c3 = c1;

while (*c3)
c3++;

while (*c3++ = *c2++)
;
return c1;
}

here a test program:

#include <stdio.h>
void main()
{
char buf[100];
sprintf(buf, "hello");
char * str = " folks!";
printf("%s\n", concat(buf, str));
}

He just missed the const reuqirement for the first argument. But this is an OP fault (i.e. David's prototype is better). :)

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]

[CPallini]

sampath-padamatinti wrote:

I am unable to solve this one. Please give me the approach

David did, here. However, if you're stuck with the const requirement on the first argument, then:

char * concat(const char * c1, const char * c2)
{
size_t size[2] = {strlen(c1), strlen(c2)};
char * c = new char[size[0]+size[1]+1];
strcpy(c, c1);
strcpy(c+size[0],c2);
return c;
}

if you can't use strlen and/or strcpy:

char * concat(const char * c1, const char * c2)
{
size_t len=0;
char * c;
const char *p;
p = c1;
while (*p++) len++;
p = c2;
while (*p++) len++;
c = new char[len+1];
p=c1;
while (*c=*p++) c++;
p=c2;
while (*c++=*p++) ;
return (c-len-1);
}

Test program:

void main()
{
const char * str1 = "hello, ";
const char * str2 = "folks!";
char * result = concat(str1, str2);
printf("%s\n", result);
delete [] result;
}

:)

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]

[Lost User]

:laugh: :laugh:

MVP 2010 - are they mad?

[CPallini]

What's that funny, Rick? :)

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]

[Lost User]

That code is too dangerous for words. As long as the length of buf is greater than the combined length of c1 and c2 plus a null terminator it will work. But as soon as the result overflows all hell breaks loose. This maybe OK for a skilled developer such as yourself, but I would not suggest it as a solution for a newbie.

MVP 2010 - are they mad?

[CPallini]

The code provide what provides strcat, no less no more. I know strcat is a very dangerous function... :rolleyes: God once said: "The newbies should learn C pointers or go to Hell managed". :-D Moreover, since that was an interview question (see [^]), the OP was expected to have such a skill. :) BTW My 5 for the 'skilled developer'. :laugh:

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]

[Lost User]

:thumbsup:

MVP 2010 - are they mad?

[Saravanan Sundaresan]

Here is the function as per your requirement:

char* concat(const char* c1, const char* c2)
{
int size1 = 0, size2 = 0;

for (int i = 0; c1\[i\] != '\\0'; i++)
	size1++;
for (int j = 0; c2\[j\] != '\\0'; j++)
	size2++;

char\* result = new char\[size1+size2\];

for(int i=0; i<size1; i++)
{
	result\[i\] = c1\[i\];
}

for(int j = 0; j <= size2; j++)
{
	result\[size1+j\] = c2\[j\];
}

result\[size1+size2\] = 0;

return result;

}