Question

I’m learning C language…

I want to write a function that concatenates two strings.
I wrote a function but it doesn’t work;
it doesn’t give any error in compilation,
but while running it doesn’t do anything.

Here is my code:

char* str_sum(char* s1, char* s2){
    int j = strlen(s1);
    int i=0;
    while(s2[i]){
            s1[j]=s2[i];
            j++;
            i++;
    }
    return s1;
}
int main(){
    char* s1, s2;
    s1 = "Joe";
    s2 = "Black";
    printf("%s\n",sum_str(s1,s2));
    return 0;
}

Answer 1

Your function could look like this:

char* sum_str(char* s1, char* s2)
{
    int lenS1 = strlen(s1);
    int lenS2 = strlen(s2);
    char* newString = malloc((lenS1 + lenS2 + 1) * sizeof(char));
    int i = 0;
    while(i < lenS1)
    {
        newString[i] = s1[i];
        i++;
    }
    while(i < lenS2 + lenS1)
    {
        newString[i] = s2[i - lenS1];
        i++;
    }
    newString[i] = '\0';
    return newString;
}

Note that this function allocates new string which means you should free this data when you finish with this string. Also note that terminating character ('\0') is stored at the end of this char array so that printf can “print” it properly.

Here’s main:

int main()
{
  char *s1, *s2, *s3;
  s1 = "Joe";
  s2 = " Black";
  s3 = sum_str(s1,s2);
  printf("%s\n", s3);
  free(s3);
  return 0;
}

Output: Joe Black

Note that I have declared variables s1, s2 and s3 like this: char *s1, *s2, *s3;. If I write it like this: char *s1, s2, s3; then variables s2 and s3 are no longer arrays of characters but only characters.

Also note that this program:

  char *s1 = "Joe";
  s1[0] = 'X';
  printf("%s\n", s1);

will crash since it tries to change constant string literal "Joe". s1 is pointer to first character of this literal in this case.

But this program will work fine and its output will be Xoe:

  char s1[] = "Joe";
  s1[0] = 'X';
  printf("%s\n", s1);

s1 is an array initialized with string "Joe" so it’s OK to change it.