How do I concatenate 2 pointer strings correctly?

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Hi

I have tried, see the code below, to concatenate 2 pointer strings with the strcat() function. It didn't work.

// Jones, Bradley L.; Peter Aitken; Dean Miller. C Programming in One Hour a Day, Sams Teach Yourself (p. 400). Pearson Education. Kindle Edition.
//
// 7. ON YOUR OWN: Write a function that accepts two strings. Use the malloc() function to allocate enough memory to hold
// the two strings after they have been concatenated (linked). Return a pointer to this new string. For example, if you pass
// "Hello" and "World!", the function returns a pointer to "Hello World!". Having the concatenated value be the third string
// is easiest. (You might use your answers from exercises 5 and 6.)

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

  int main( void )
  {
    char *A = "Hello";
    char *A2 = " ";
    char *A3 = "World!";

    strcat(A,A2);
    printf("\n%s\n",A);

    return 0;
 }

How do I concatenate 2 pointer strings correctly?

What I have tried:

I have tried to change from

printf("\n%s\n", A);

to

printf("\n%s\n",A);.

.

I have also tried with:

printf("\n%s\n", strcat(A,A2));

.

Posted 9-Oct-21 1:39am

Mieczyslaw1683

Updated 9-Oct-21 3:16am

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2 solutions

Solution 2

The best way is to do use malloc to allocate the memory to copy it into, then use memcopy - as Richard says, trying to change a constant string can crash your app.

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main()
    {
    printf("Hello World\n");
    char *A = "Hello";
    char *B = " World";
    int lenA = strlen(A);
    int lenB = strlen(B);
    char *A3 = (char*) malloc(lenA + lenB + 1);
    memcpy(A3, A, lenA);
    memcpy(A3 + lenA, B, lenB);
    A3[lenA + lenB] = '\0';
    printf("\n%s\n",A3);

    return 0;
    }

malloc provides a way to allocate memory of a variable size - so it's easier to write a function that can be called repeatedly to do the copy to fresh memory:

char* AppendStrings(const char *A, const char*B)
    {
    int lenA = strlen(A);
    int lenB = strlen(B);
    char *C = (char*) malloc(lenA + lenB + 1);
    memcpy(C, A, lenA);
    memcpy(C + lenA, B, lenB);
    C[lenA + lenB] = '\0';    
    return C;
    }

Then you can call it with whatever you want:

char * C  = AppendStrings(A, B);
printf("\n%s\n",C);
char * D  = AppendStrings("Hello World!\n", "This is a test");
printf("\n%s\n",D);

Posted 9-Oct-21 3:16am

OriginalGriff

Comments

Rick York 9-Oct-21 12:03pm

You can also use calloc which will zero the allocated memory so the terminating null character is there automatically.

OriginalGriff 9-Oct-21 12:19pm

I tend to forget calloc - it's been a long time since I wrote any non-embedded C code ... so I haven't called malloc or calloc in the real world for decades! :laugh:

KarstenK 9-Oct-21 12:45pm

It has to be mentioned that the allocated memory has to be freed when done, else it is leaking.

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Richard MacCutchan · Accepted Answer · 2021-10-09T02:34:00

You are concatenating a single space to A so it probably appears not to work (you forgot A3). However, trying to write into a constant string is never guaranteed and you should not do it. You should create a new char buffer and copy the text into that. Something like:

C++

char *A = "Hello";
char *A2 = " ";
char *A3 = "World!";

char buffer[32]; // make sure this is big enough for the final text
strcpy(buffer, A);    // copy the first string into the buffer
strcat(buffer, A2);   // concatenate the space
strcat(buffer, A3);   //     and the final string
printf("\n%s\n", buffer);