I'm currently trying to run Export to PDF script, however when I try running it in DOORS ver 9.6.1, I get a couple of errors.
Line 8: char c = str[i]; contains a syntax error
Any thoughts on how I can resolve this issue?
string makeCaption(Buffer& str)
{
setempty(tempBuf);
int i = 0;
{
for(i = 0; i < length(str); ++i)
char c = str[i];
if('\n' != c) && '\\' != c)
{
tempBuf += c;
}
}
escapeSpecialLaTeXCharacters(tempBuf);
return stringOf(tempBuf);
}
Seems to me like copy/paste problems. When you compare your code with the original you might notice that you moved line 6 with the sole { one line up. If you put it back where it belongs i.e. after the line "for(...)", the code works
I'm looking to do something similar to this how to get integer variable name and its value from Expr* in clang using the RecursiveASTVisitor
The goal is to first retrieve all assignment operations then perform my own checks on them, to do taint analysis.
I've overridden the VisitBinaryOperator as such
bool VisitBinaryOperator (BinaryOperator *bOp) {
if ( !bOP->isAssignmentOp() ) {
return true;
}
Expr *LHSexpr = bOp->getLHS();
Expr *RHSexpr = bOp->getRHS();
LHSexpr->dump();
RHSexpr->dump();
}
This RecursiveASTVisitor is being run on Objective C codes, so I do not know what the LHS or RHS type will evaluate to (could even be a function on the RHS?)
Would it be possible to get the text representation of what is on the LHS/RHS out from clang in order to perform regex expression on them??
Sorry, I found something similar that works for this particular case.
Solution:
bool VisitBinaryOperator (BinaryOperator *bOp) {
if ( !bOP->isAssignmentOp() ) {
return true;
}
Expr *LHSexpr = bOp->getLHS();
Expr *RHSexpr = bOp->getRHS();
std::string LHS_string = convertExpressionToString(LHSexpr);
std::string RHS_string = convertExpressionToString(RHSexpr);
return true;
}
std::string convertExpressionToString(Expr *E) {
SourceManager &SM = Context->getSourceManager();
clang::LangOptions lopt;
SourceLocation startLoc = E->getLocStart();
SourceLocation _endLoc = E->getLocEnd();
SourceLocation endLoc = clang::Lexer::getLocForEndOfToken(_endLoc, 0, SM, lopt);
return std::string(SM.getCharacterData(startLoc), SM.getCharacterData(endLoc) - SM.getCharacterData(startLoc));
}
Only thing I'm not very sure about is why _endLoc is required to compute endLoc and how is the Lexer actually working.
EDIT:
Link to the post I found help Getting the source behind clang's AST
This is my function. It's working absolutely fine; I just can't get one more thing working.
Instead of the static fopen paths, I need the user to write the path for the files. I tried several things but I can't get it working. Please help
int FileToFile() {
FILE *fp;
FILE *fp_write;
char line[128];
int max=0;
int countFor=0;
int countWhile=0;
int countDo = 0;
fp = fopen("d:\\text.txt", "r+");
fp_write = fopen("d:\\results.txt", "w+");
if (!fp) {
perror("Greshka");
}
else {
while (fgets(line, sizeof line, fp) != NULL) {
countFor = 0;
countWhile = 0;
countDo = 0;
fputs(line, stdout);
if (line[strlen(line)-1] = "\n") if (max < (strlen(line) -1)) max = strlen(line) -1;
else if (max < strlen(line)) max = strlen(line);
char *tmp = line;
while (tmp = strstr(tmp, "for")){
countFor++;
tmp++;
}
tmp = line;
while (tmp = strstr(tmp, "while")){
countWhile++;
tmp++;
}
tmp = line;
while (tmp = strstr(tmp, "do")){
countDo++;
tmp++;
}
fprintf(fp_write, "Na tozi red operatora for go ima: %d pyti\n", countFor);
fprintf(fp_write, "Na tozi red operatora for/while go ima: %d pyti\n", countWhile - countDo);
fprintf(fp_write, "Na tozi red operatora do go ima: %d pyti\n", countDo);
}
fprintf(fp_write, "Maximalen broi simvoli e:%d\n", max);
fclose(fp_write);
fclose(fp);
}
}
Have a look at argc and argv. They are used for command-line arguments passed to a program. This requires that your main function be revised as follows:
int main(int argc, char *argv[])
The argc is an integer that represents the number of command-like arguments, and argv is an array of char* that contain the arguments themselves. Note that for both, the program name itself counts as an argument.
So if you invoke your program like this:
myprog c:\temp
Then argc will be 2, argv[0] will be myprog, and argv[1] will be c:\temp. Now you can just pass the strings to your function. If you pass more arguments, they will be argv[2], etc.
Keep in mind if your path contains spaces, you must enclose it in double quotes for it to be considered one argument, because space is used as a delimiter:
myprog "c:\path with spaces"
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.
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