Write strcat() function with pointers

Question

I am new with pointers on C and I am trying to write a function like strcat() but without using it. I developed the following function:

char cat(char *a, char *b) {
 int i=0,cont=0,h=strlen(a)+strlen(b);
 char c[h]; //new string containing the 2 strings (a and b)

 for(i;i<strlen(a);++i) {
  c[i] = *(a+i); //now c contains a      
 }

 int j = i;

 for(j;j<strlen(b);++j) {
  c[j] = *(b+cont); //now c contains a + b
  cont++;       
 }

 return c; // I return c  
}

And this is how I call the function:

  printf("\Concatenazione: %c", cat(A,B));

It is now working because the final result is a weird character. How could I fix the function? Here there's the full main.

Listen to your compiler's warnings; gcc says, return makes integer from pointer without a cast. You cannot return a character array from a function that returns char. — user4815162342
– user4815162342, Commented Dec 10, 2013 at 13:36
I do not understand why 4 of 5 answers are using malloc. strcat copies source to dest and returns dest. malloc should not apply here. — Duck
– Duck, Commented Dec 10, 2013 at 14:09
@Duck i agree, the char array should be declared before calling the function and should be passed to the function by a pointer. — H_squared
– H_squared, Commented Dec 10, 2013 at 14:42

score 2 · Accepted Answer · 2013-12-10 13:56:25Z

2

char * strcat(char *dest, const char *src)
{
    int i;
    int j;

    for (i = 0; dest[i] != '\0'; i++);
    for (j = 0; src[j] != '\0'; j++) {
        dest[i+j] = src[j];
    }

    dest[i+j] = '\0';

    return dest;
}

edited Dec 10, 2013 at 13:56

answered Dec 10, 2013 at 13:38

user3084858

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2 Comments

drew_w Over a year ago

If you are passing dest in, then you don't need to return it. But the idea of this question (in my understanding) was to concatenate two strings into a new string, which is not what this does. This actually edits the first string passed into the routine!

Duck Over a year ago

@drew_w, maybe. OTOH OP said he wanted to recreate strcat. OTOH his implementation either purposely created something else (a new string) or his attempt just went awry and most of the answers followed that objective. This answer comes closest to what strcat does and the behavior expected.

Sergey Kalinichenko · Accepted Answer · 2013-12-10 14:14:05Z

2

From your implementation it appears that your version of strcat is not compatible with the standard one, because you are looking to allocate memory for the result, rather than expecting the caller to provide you with enough memory to fit the result of concatenation.

There are several issues with your code:

You need to return char*, not char
You need to allocate memory dynamically with malloc; you cannot return a locally allocated array.
You need to add 1 for the null terminator
You need to write the null terminator into the result
You can take both parameters as const char*
You can simplify your function by using pointers instead of indexes, but that part is optional.

Here is how you can do the fixes:

char *cat(const char *a, const char *b) {
    int i=0,cont=0,h=strlen(a)+strlen(b);
    char *c = malloc(h+1);
    // your implementation goes here
    c[cont] = '\0';
    return c;
}

edited Dec 10, 2013 at 14:14

answered Dec 10, 2013 at 13:37

Sergey Kalinichenko

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2 Comments

Sergey Kalinichenko Over a year ago

@Marco Thanks! (Feel free to edit when you see such stupid mistakes).

Alberto Rossi Over a year ago

Yeah, I have to use C

drew_w · Accepted Answer · 2013-12-10 16:25:38Z

You are returning a POINTER to the string, not the actual string itself. You need to change the return type to something like "char *" (or something equivalent). You also need to make sure to null terminate the string (append a '\0') for it to print correctly.

Taking my own advice (and also finding the other bug, which is the fact that the second for loop isn't looping over the correct indices), you end up with the following program:

#include <stdio.h>

char *cat(char *a, char *b) {
  int i = 0, j = 0;
  int cont = 0;
  int h = strlen(a) + strlen(b) + 1;

  char *result = (char*)malloc(h * sizeof(char));

  for(i = 0; i < strlen(a); i++) {
    result[i] = a[i];
  }

  for(j = i; j < strlen(b)+ strlen(a); j++) {
    result[j] = b[cont++];
  }

  // append null character
  result[h - 1] = '\0';
  return result;
}

int main() {
   const char *firstString = "Test First String. ";
   const char *secondString = "Another String Here.";
   char *combined = cat(firstString, secondString);

   printf("%s", combined);

   free(combined);
   return 0;
}

Marco · Accepted Answer · 2013-12-10 13:35:34Z

1

c is a local variable. It only exists inside the function cat. You should use malloc.

instead of

char c[h];

use

char *c = malloc(h);

Also, you should add the null byte at the end. Remember, the strings in C are null-ended.

h = strlen(a) + strlen(b) + 1;

and at the end:

c[h - 1] = '\0';

The signature of cat should be char *cat(char *a, char *b);

answered Dec 10, 2013 at 13:35

Marco

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Comments

Abhineet · Accepted Answer · 2013-12-10 13:49:28Z

You will get an error of

expected constant expression

for the code line char c[h];. Instead you should be using malloc to allocate any dynamic memory at run-time like::

char* c ;
c = malloc( h + 1 ) ; // +1 for the terminating null char
// do stuff
free( c ) ;

Your corrected code::

#include<stdio.h>
#include<conio.h>
#include<string.h>
#include <stdlib.h>

char* cat(char *a, char *b) {
 int i=0,cont=0,h=strlen(a)+strlen(b), j;
 char *c;

 c = malloc( h+1 ) ;
 for(i;i<strlen(a);++i) {
  c[i] = *(a+i);      
 }

 j = 0 ;

 for(j;j<strlen(b);++j) {
  c[i] = *(b+cont);
  i++ ;
  cont++;      
 }

 c[i] = 0 ;

 return c;    
}

int main() {

  char A[1000],B[1000];
  char * a ;

  printf("Inserisci la stringa 1: \n");
  gets(A);
  printf("Inserisci la stringa 2: \n");
  gets(B);
  a = cat(A,B) ;
  printf("\nConcatenazione: %s", a);

  free(a) ;

 getch();
 return 0;    
}

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