Question

I have to make a function which concatenates two strings but I have to add a ‘\n’ after the first word. I figured everything out and for some reason it doesn’t print out anything. Any ideas? It probably has to do something with the pointers. I just can’t get my head around them. Here’s the code.

char *function(char *s1, char *s2){
    char *newStr;
    int size;
    size = strlen(s1) + strlen(s2);

    newStr = (char *)malloc((size+1)*sizeof(char));

    while(*s1!= '\0'){
        *newStr = *s1;
        newStr++;
        s1++;
    }
    *newStr = '\n';
    newStr++;
    while(*s2 != '\0'){
        *newStr = *s2;
        newStr++;
        s2++;
    }
    *newStr = '\0';
return newStr;
}

int main (int argc, const char * argv[]) {
    char *str1 = "Hello";
    char *str2 = "World";

    printf("%s",function(str1, str2));

     return 0;
}

So as a result I should get:

Hello
World

but I’m not getting anything back.

Answer 1

You are returning a pointer to the end of the buffer rather than a pointer to the start of the buffer. Look at the last two lines of the function:

*newStr = '\0';
return newStr;

Clearly this returns a pointer to the null char, i.e. the empty string.

Solve the problem by introducing a temporary pointer which you will use to step through the output buffer. Then you can return the pointer to the beginning of the output buffer.

char *function(char *s1, char *s2){
    int size = strlen(s1) + strlen(s2) + 2;//one for '\n', one for '\0'
    char *result = malloc(size);
    char *p = result;

    while(*s1 != '\0'){
        *p = *s1;
        p++;
        s1++;
    }
    *p = '\n';
    p++;
    while(*s2 != '\0'){
        *p = *s2;
        p++;
        s2++;
    }
    *p = '\0';

    return result;
}

You also need to allocate an extra char for the \n, as shown above. Finally, your calling code never frees the memory allocated by function.