pass array to memset

[sef0]

Hi,

#include <stdio.h>
#include <string.h>

char myArray[] =
"\x31\x32\x31\x32\x31\x32x31\"
"\x32\x32\x43\x54\x65\x78\x6e\x61";

int main(void)
{
    char myMem[225];
    memset(myMem, &myArray, 15);
    return(0); 
}

I'd like to pass the myArray to memset's 2nd parameter, that is, replicate myArray 15 times inside myMem[]. Is it doable?

[quzah]

You mean like strcpy? Or like memmove?


Quzah.

[sef0]

You mean like strcpy? Or like memmove?
Quzah.

Neither. I'm going to use

memcpy(&myMem[225-strlen(myArray)],myMem,strlen(myArray));

But the problem I encounter is on the memset();
I don't know how to pass an array as second argument.

memset(myMem, ???myArray???, 15);

[quzah]

You mean like strcpy or memmove?


Quzah.

[matsp]

memset is not able to do this. As far as I'm aware, there is no generic function in C that would copy a block of memory x number of times.

You can of course write a function that takes the relevant char * arguments and duplicates it the necessary number of times. You could for example use a loop using memcpy().

By the way, would it not be easier to write your string as "121212122CTNA", rather than the hex notation? [If you think that is actually hiding some sort of password or such, I can guarantee that looking at the memory in a debugger or using the "strings" command in Linux/Unix will show the same thing whether you wrote it in clear-text or as hex-codes, so it's only (but not very) meaningful to be cryptic about it in the source file, and it only makes it hard for someone who don't know the ascii table or don't understand hex - and those people would probably barely recognize a piece of C code as something readable anyways!].

--
Mats

[itCbitC]

As noted call memcpy(), iteratively or recursively, "x" number of times to populate the destination array.