C char* array Value assignment index

Question

While creating strjoin I noticed a strange phenomenon. When outputting a value strs in function strjoin()

These results were printed

123

Why?

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

char    *strjoin(int size, char **strs, char *sep)
{
    char *new;

    if (!(new = malloc(sizeof(char) * size)))
        return (NULL);

    printf("%s", strs[1]);
    return (NULL);
}

#include <unistd.h>

int     main(void)
{
    char* b[4] = {"123", "456", "789", "245"};
    char *p;
    int i = 0;
    int size = 5;
    char a[4] = {',', 'a', 'b', '\0'};

    p = *b;

    strjoin(5, &p, a);
}

Answer 1

In order for it to be valid to use the value referenced by str[1] , str must point to an array of at least two char* elements. In your demonstration, it points to p , which is a single char* . Your program therefore invokes Undefined Behaviour , and thus your program is invalid.

It's unclear what you are trying to accomplish. Perhaps you wanted

char **p = &( b[0] );
strjoin(4, p, a)

Keep in mind that an array used where a pointer is expected degenerates into a pointer to its first elements, so the above is equivalent to the following:

char **p = b;
strjoin(4, p, a)

or simply

strjoin(4, b, a)

(The first parameter to strjoin isn't being used in your demonstration, but I imagine it's expected to be the number of elements in *str , so I have adjusted it the argument accordingly.)