On Fri, 2013-10-04 at 18:13 +0000, Rafael Ratacheski wrote:
>
> Your solution resolve my problem partially, I have success to print
> the value but my real objective is save the value in a buffer, and
> when i try do this, with your method, i don't have success. The printf
> line is just to verify the saved value.
>
> About the memory allocated, i think this resolve
> size_t test = 64;
> int *sp = malloc(test);
> memcpy(sp, vars->val.string, vars->val_len);
Kind of. You are allocating 64 bytes to store a 6-byte value.
> int *teste = htonl(*sp);
Why are you insisting on byteswapping the first four bytes of the value?
Why are you insisting on storing them in a pointer??? (not in the
pointed to value, in the pointer!)
> printf("value #%d is a string: %x \n", count++, teste);
Here you are only trying to print the first four bytes of the value. I
suppose that is the reason for the last two bytes to not be visible.
> free(sp);
void printbuf(const void* buf, size_t buflen)
{
const unsigned char* bp = (const unsigned char*)buf;
while (buflen--)
printf("%02x", *bp++);
}
...
printfbuf(vars->val.string, vars->val_len);
puts("");
char *sp = malloc(vars->val_len);
memcpy(sp, vars->val.string, vars->val_len);
printbuf(sp, vars->val_len);
puts("");
free(sp);
/MF
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