Adding a string to another string

Question

Exercise:

Write a function that returns a CH character string after adding a CH1 string at the end of a CH2 string given as a parameter.

Here is my code:

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

char *ajout(char ch1[], char ch2[]);

int main() {
    char ch1[] = "hello";
    char ch2[] = "ooo";
    char *ch = ajout(ch1, ch2);
    printf("%s",ch);
    return 0;
}

char *ajout(char ch1[], char ch2[]) {
    char *ch;
    int nb = 0;
    for (int i = 0; ch1[i] != '\0'; i++) {
        ch[i] = ch1[i];
        nb++;
    }
    int i = 0;
    for (i; ch2[i] != '\0'; i++) {
        ch[nb] = ch2[i];
        nb++;
    }
    return ch;
}

The expected result after program execution: helloooo

char *ch; has no memory allocated (in the function ajout). Try ch = malloc(1 + strlen(ch1) + strlen(ch2)); And you should add a nul terminator to the concatenated string. — Weather Vane
– Weather Vane, Commented Dec 20, 2019 at 20:20
Does this answer your question? How do I concatenate two strings in C? — Alexander van Oostenrijk
– Alexander van Oostenrijk, Commented Dec 20, 2019 at 20:23
thank you @weathervane that's what i do :` int c = malloc(1 + strlen(ch1) + strlen(ch2)); char ch[c]; *ch=ajout(ch1,ch2);` but it's still not woking,what's the wrong with my code ? — Yassel
– Yassel, Commented Dec 20, 2019 at 20:34
Post any update in the question (full implementation) so that we can revision it. — Roberto Caboni
– Roberto Caboni, Commented Dec 20, 2019 at 20:36

score 3 · Accepted Answer · 2019-12-20 21:51:22Z

First, a string in C is just a pointer to an array of char that is terminated by the first null character. There is no string concatenation operator in C.

Also, in order to create a new array in C - you need to allocate memory for it using malloc. However, in the ajout function, you don't allocate any memory for ch. Solution:

const size_t len1 = strlen(ch1); //get ch1 length
const size_t len2 = strlen(ch2); //get ch2 length
char *ch = malloc(len1 + len2 + 1); //create ch with size len1+len2+1 (+1 
                                    //for null-terminator)

I see you copy each char of ch1 and ch2 into ch one by one. A better solution would be to copy ch1 and ch2 using memcpy. Also, after allocation memory using malloc, you need to deallocate the string using free:

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
char* ajout(const char *ch1, const char *ch2)
{
    //get length of ch1, ch2
    const size_t len1 = strlen(ch1);
    const size_t len2 = strlen(ch2);

    //create new char* array to save the result 
    char *result = malloc(len1 + len2 + 1); // +1 for the null-terminator
    // in real code you would check for errors in malloc here

    //copy ch1, ch2 to result
    memcpy(result, ch1, len1); 
    memcpy(result + len1, ch2, len2 + 1); // +1 to copy the null-terminator
    return result;
}

int main()
{
    char ch1[] = "hello";
    char ch2[] = "ooo";
    char *ch = ajout(ch1,ch2);
    printf("%s",ch);
    free(ch); // deallocate the string
    return 0;
}

output: helloooo

You can read more about memcpy here and about malloc here.

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