使用memcpy将两个char *复制到** buff-C语言

Question

I got two char pointer:

char *a="A";
char *b="B";

And a pointer to pointer buffer:

char **buf = malloc(sizeof(char*)*2);

And I want use memcpy to copy two variables to buf:

memcpy(*buf, &a, sizeof(char*));
memcpy(*buf, &b, sizeof(char*));

but it replace first variable..

how can I copy two?

Answer 1

What is it you actually want to do?

With

char **buf = malloc(sizeof(char*)*2);

memcpy(*buf, &a, sizeof(char*));
memcpy(*buf, &b, sizeof(char*));

unless you omitted some initialisation in between, you get undefined behaviour. The contents of the malloc ed memory is unspecified, so when *buf is interpreted as a void* in memcpy , that almost certainly doesn't yield a valid pointer, and the probability that it is a null pointer is not negligible.

If you just want buf to contain the two pointers, after the malloc

buf[0] = a;
buf[1] = b;

is the simplest and cleanest solution, but

memcpy(buf, &a, sizeof a);
memcpy(buf + 1, &b sizeof b);

would also be valid (also with &buf[1] instead of buf + 1 .

If you want to concatenate the strings a and b point to, you're following a completely wrong approach. You'd need a char* pointing to a sufficiently large area to hold the result (including the 0-terminator) and the simplest way would be using strcat .

Answer 2

I tried your program with my own test code: I hope this helps:

#define MAX_LENGTH 10
#define ARRAY_LENGTH 2
int main(int argc, char *argv[])
{
    char *a = "A";
    char *b = "B";
    char **buf = (char **)malloc(sizeof(char*)*ARRAY_LENGTH);
    *buf = (char *)malloc(sizeof(char) * MAX_LENGTH);
    *(buf+1) = (char *)malloc(sizeof(char) * MAX_LENGTH);
    memset(buf[0],0,10);
    memset(buf[1],0,10);
    memcpy(buf[0],a,strlen(a));
    memcpy(buf[1],b,strlen(b));
    printf("%s %s",buf[0],buf[1]);
    free(buf[0]);
    free(buf[1]);
        free(buf)
    return 0;
}