urgent help:string concat

jack · #1 11-Apr-2004, 02:49

hi below there is a simple concatenation program. but i'm getting segmentation fault in it. the problem is that i'm tring to concatenate an integer value to the string but in vain. please help me.
thnx

CPP / C++ / C Code:

#include <stdio.h>

int main()
{
	char *s1,*s2, *s3, *all, *result, *thecode;
int x=567;
	s1 = "Hello ";
	s2 = "world! ";
	s3 = (char *)x;

	all= (char *)calloc(strlen(s1) + strlen(s2) + strlen(s3), sizeof(char));

	strcat (all,s1);
	strcat (all,s2);
	strcat (all,s3);


	printf ("The string is = %s\n",all);
}

dsmith · #2 11-Apr-2004, 07:28

Hello Jack. I don't think that strcat is your problem as much as this:

CPP / C++ / C Code:

s3 = (char *)x;

This doesn't do what you think it does. You can not typecast a string from an integer. This is going to set the address of s3 to memory location 576.

Also, on strcat, you probably wouldn't get the result you wanted either, because it is going to entirely overwrite the string. I have modified your code to get the results you wanted:

CPP / C++ / C Code:

#include <stdio.h>

int main()
{
	char *s1,*s2, *s3, *all, *result, *thecode;
int x=567;
	s1 = "Hello ";
	s2 = "world! ";
	s3 = itoa(x);

	all= (char *)calloc(strlen(s1) + strlen(s2) + strlen(s3), sizeof(char));

	strcat (all,s1);
	strcpy (all,s2);
	strcpy(all,s3);


	printf ("The string is = %s\n",all);
}

Unfortunately, most compilers don't have itoa (which is the opposite of atoi). If you are so unfortunate, here is a routine that I wrote quite some time ago to convert numbers to a char*

CPP / C++ / C Code:

char *numasc(int num)
{
	char	string[30],
			*ret;
	int		digit,
			left,
			pos,
			negative = 0;


	if(num == 0){
		ret = new char[2];
		*ret = '0';
		*(ret+1) = 0;
		return ret;
  	}
	string[29] = 0;
	left = num;
	pos = 28;
	if(left < 0){
		left = abs(left);
		negative = 1;
	}

	while((left) && (pos)){
		digit = left % 10;
		left = (int) left/10;

		string[pos] = digit+48;
		pos--;
	}
	if(negative){
		string[pos] = '-';
		pos--;
  	}
    ret = new char[30-pos];
	strcpy(ret,&string[pos+1]);

	return(ret);
}

Hope this helps,
d

WaltP · #3 11-Apr-2004, 08:46

D, the problems I noticed with your routine is that the string[] array is a local buffer that goes out of scope as soon as the routine returns, therefore the data technically does not exist upon return.

Instead you can use sprintf() to convert the number to a string:

CPP / C++ / C Code:

  char *s1,*s2, s3[10], *all, *result, *thecode;  // You must allocate space for s3
  int x=567;
  s1 = "Hello ";
  s2 = "world! ";
  sprintf(s3, "%d", x);

dsmith · #4 11-Apr-2004, 09:03

Quote:

Originally Posted by WaltP

D, the problems I noticed with your routine is that the string[] array is a local buffer that goes out of scope as soon as the routine returns, therefore the data technically does not exist upon return.

You are absolutely correct. Like I said, I wrote this a while ago and have always wanted to fix it to be more efficient.

I keep forgetting about sprintf. You could avoid strcat all together with sprintf.

CPP / C++ / C Code:

#include <stdio.h>

int main()
{
  char *s1,*s2,*all, *result, *thecode;
int x=567;
  s1 = "Hello";
  s2 = "world!";

  all= (char *)calloc(strlen(s1) + strlen(s2) + 10, sizeof(char));

  sprintf(all,"%s %s %d",s1,s2,x);
  printf ("The string is = %s\n",all);
}

I like this much better. You can do all of your formatting with sprintf and get a single string output. Notice that I removed the spaces from s1 and s2 and included them in the sprintf statement.

Thanks Walt.

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