I have a bunch of test cases in an XML file named : blah_blah_blah_blah_number
The test cases have the numbers all messed up like:
blah_3
blah_1
blah_7
....
I need to re-number them. So that the first one gets renumbered 1, the second 2.. and so on. I want to build a macro for this but I don't know how to start. I need some sort of search function that can go to the pattern I give it, and then it substitutes the number with a count that I keep in some variable. I'm not proficient at all with Slick-C, and would like to get this done quickly :\
Any help appreciated,
Ted
For a more timely answer, you might consider the SlickEdit forums on slickedit.com, but I'll try here.
I would load the file in a buffer, and create a macro along the following lines:
Create a while loop incrementing a counter
In the while loop, use search with regular expressions to find the next occurrence
When the next occurrence is found, use search_replace to replace the string found with a new string composed using the counter variable, and to repeat the previous search
When nothing is to be found any more, terminate the loop
Without testing it on an XML file, this should give you a start for your function (assuming you know how to create a macro file and load it into SE), a quick test showed it to work on a plain text file, no guarantees, however:
/* Demo for StackOverflow question 14205293 */
_command my_renumber() name_info(','VSARG2_MARK|VSARG2_REQUIRES_EDITORCTL)
{
int not_found = 0; /* boolean to terminate loop */
int iLoop = 0; /* Counter to renumber items */
/* Use search initially; if that call doen't find an item, we're done. */
if (search('blah_:i', 'R') != 0)
{
not_found = 1;
}
while (!not_found)
{
if (search_replace('blah_' :+ iLoop, 'R') == STRING_NOT_FOUND_RC)
{
not_found = 1;
}
iLoop++;
}
}
Related
I was told that I should be using while(fin) instead of while(!fin.eof()) when reading a file.
What exactly is the difference?
Edit: I do know that while(fin) actually checks the stream object and that when it becomes NULL, the loop breaks and it covers eof and fail flags.
But my course teacher says that fin.eof() is better so I need to understand the fundamental operation that's going on here.
Which one is the right practice?
Note: This is not a duplicate, I need assistance in Turbo C++ and with binary files.
I'm basically trying to read a file using a class object.
First of all I am assuming fin is your fstream object. In which case your teacher would not have told you to use while(fin.eof()) for reading from file. She would have told to use while(!fin.eof()).
Let me explain. eof() is a member of the fstream class which returns a true or false value depending on whether the End Of File (eof) of the file you are reading has been reached. Thus while eof() function returns 0 it means the end of file has not been reached and loop continues to execute, but when eof() returns 1 the end of the file has been reached and the loop exits.
while(fin) loop is entered because fin actually returns the value of an error flag variable inside the class object fin whose value is set to 0 when any function like read or write or open fails. Thus the loop works as long as the read function inside the loop works.
Personally I would not suggest either of them.
I would suggest
//assume a class abc.
abc ob;
While(fin.read((char*)&ob, sizeof(ob)))
{}
Or
While(fin.getline(parameters))
{}
This loop reads the file record inside the loop condition and if nothing was read due to the end of file being reached, the loop is exited.
The problem with while(!fin.eof()) is that it returns 1 if the end of file has been reached. End of file is actually a character that is put at the end of the file. So when the read function inside the loop reads this character and sets a variable eof to 1. All the function actually does is return this value.
Thus works fine when you are reading lines in words but when you are reading successive records of a class from a file, this method will fail.
Consider
clas abc
{}a;
Fstream fin("file");
While(!fin.eof())
{
fin.read((char*)&a,sizeof(a));
a.display(); // display is a member function which displays the info }
Thus displays the last record twice. This is because the end of file character is the character after the last byte of the last record. When the last is read the file pointer is at the eof byte but hasn't read it yet. So it will enter the loop again but this time the eof char is read but the read function fails. The values already in the variable a, that is the previous records will be displayed again.
One good method is to do something like this:
while ( instream.read(...) && !instream.eof() ) { //Reading a binary file
Statement1;
Statement2;
}
or in case of a text file:
while ( (ch = instream.get()) && !instream.eof() ) { //To read a single character
Statement1;
Statement2;
}
Here, the object is being read within the while loop's condition statement and then the value of eof flag is being tested.
This wouldn't result in undesired outputs.
Here we are checking the status of the actual I/O operation and the eof together. You may also check for the fail flag.
I would like to point out that according to #RetiredNinja, we may only check for the I/O operation.
That is:
while ( instream.read(...) ) { //Reading a binary file
Statement1;
Statement2;
}
A quick and easy workaround that worked for me to avoid any problems when using eof is to check for it after the first reading and not as a condition of the while loop itself. Something like this:
while (true) // no conditions
{
filein >> string; // an example reading, could be any kind of file reading instruction
if (filein.eof()) break; // break the while loop if eof was reached
// the rest of the code
}
This is a very basic question but please bear with me!
I got this code in a question as part of a quiz I was doing earlier and just didn't know if I might be missing something. I typed it into the editor and it would not run and it does appear to be incomplete. Had it been if (k) it would have made more sense.
But, as I have heard that you can leave out components of a for loop, I was just wondering if there is any time you would see the likes of for (k)?
int k = 0;
for (k) {
printf ("hello");
}
for(int k; ;)
/*this is the correct syntax of a for loop without conditional statement and incrementation/decrementation statement*\
Remember,those semi-colons within the paranthesis is important(without that the program wouldn't compile).
Now,to answer some of the questions you asked me in the earlier answer-
for(int k; ;)
{
printf("infinite loop");
}
When will this loop come to an end?
This loop will never come to an end.It is an infinite loop.It will keep printing infinite loop forever.
Is it possible to bring this loop to an end?
Yes,it is.It can be brought to an end using break statement.
for(int k; ;)
//or for( ; ; )
{
printf("infinite loop");
break;
}
Prints infinte loop only once.It will encounter the break statement and the control will move outside the loop.
Possible application.
It's used when you actually have no idea about when a loop should come to an end.
int i=0; //to take user input
for(int k; ;)
{
//accept the value of k from user.
/*You want the user to enter 1 as the input*/
if(k==1)
{
printf("entered 1,moving out of loop");
break;
}
}
What is the meaning of above loop?
- This loop keeps running until the user enters '1'.This is important in cases where you are giving the user options and the options are limited and so you don't want the user to give an invalid input.It runs until a valid input is entered(you can add more if statements with break statement).
Menu: 1)Pizza
2)Burger
3)Quit buying!
for(int k;k<10;k++)
/* this is a finite loop and this isn't suitable for the above requirement because you are not sure if the user will give the valid input within 10 iterations.*/
When k becomes 10,the control will move out irrespective of whether the user has entered a valid input or not.What if the user inputs 8 when k=9? The control will move out of the loop at k=10.As a result,your program will not work efficiently because i=8 is not an input you expected.You wanted 1,2 or 3 as input.
So,an infinite loop is used when you are not sure about how many iterations are required.You will actually be using a break statement to exit such a loop.
Is this the only option for an infinite loop?Why not while() ?Isn't while() with no condition an infinite loop?
while();// invalid in C.
//objective-C follows C-standards.
while("condition"); //valid
Some valid for loop declarations in C
for(k; ;) // infinite loop
for(; ;) // infinite loop
for(; k<0;)// valid
So,I think that sums up a small explanation.
Remember,semi-colons are important(irrespective of whether a condition is given or not).
And of course,you have other options to keep running or taking user input unless a valid input is given.But above one was just an application I could figure out to show that an infinite loop could be cool!
If you find any error or doubt,please comment.
Well,even I am not too good in C.But yeah since java is somewhat similar,I figured it out.
I've seen a similar topic on sof but its solution did not help me. This is ticking my mind and basically all i want is to have some method of accessing and modifying a value that will maintain its last changed state through out my macros in my single .ahk file.
See example below ,
~Home::Suspend
XButton1::
tog()
return
LButton::
shot()
return
var := "1"
tog(){
var *= -1
}
shot(){
If (var = "1") {
Loop, 1 {
Send {k}
Sleep 65
Send {WheelDown}
Sleep 100
Send {WheelUP}
Sleep 10
}
} Else {
Send {k}
}
}
I am aware that the above is incorrect, and i tried to use"global" in my functions but i just couldn't get my desired effect.
Using the "global" should work. Something like:
shot(){
global var
If (var = "1") {
That points the 'var' variable in the shot() function to the existing 'var' variable defined outside the function.
I had the same issue and after some trial and error I found my mistake which is the same as in the provided code:
The correct way to declare a global is before other functions
var := "1"
XButton1::
; code...
return
The code in the OPs script will hit the return first and never declare the variable
XButton1::
; code...
return ; Returns Here
var := "1" ; Incorrect Will Not Be Declared
I just wanted to provide this as an answer because while I did see this information in one of the comments, I didn't see it until after I'd already spent an additional hour figuring it out myself. As this is the answer I needed, having it as an actual prominent answer may help someone else save time.
What I did, especially since I sometimes have multiple scripts running that need to access the same variable, is to place the var in a .ini file. I also use this to preserve the variable value after a restart. The solution is somewhat slower since the data is saved to the hard disk / SSD, but it works beautifully.
Example of writing the value "S" to variable "State" in group "Finish"
IniWrite, S, C:\2Podcasts\FinishOptions.ini, Finish, State
In an other script (other AutoHotKey instance), I read the value and assign it to the variable "FinishOption".
IniRead, FinishOption, C:\2Podcasts\FinishOptions.ini, Finish, State
If you want to toggle values (True/False), you could use this. This will do an IF on the current value of the variable AND set the variable to the opposite value.
If (MyLoop := !MyLoop) ; Toggle the variable "MyLoop" True/False
{
Do something
}
Else
{
Do something else
}
return
I am trying use ANTLR to analyse a large set of code using full Java grammar. Since ANTLR needs to open all the source files and scan them, I am wondering if it can also return lines of code.
I checked API for Lexer and Parser, it seems they do not return LoC. Is it easy to instrument the grammar rule a bit to get LoC? The full Java rule is complicated, I don't really want to mess a large part of it.
If you have an existing ANTLR grammar, and want to count certain things during parsing, you could do something like this:
grammar ExistingGrammar;
// ...
#parser::members {
public int loc = 0;
}
// ...
someParserRule
: SomeLexerRule someOtherParserRule {loc++;}
;
// ...
So, whenever your oparser encounters a someParserRule, you increase the loc by one by placing {loc++;} after (or before) the rule.
So, whatever your definition of a line of code is, simply place {loc++;} in the rule to increase the counter. Be careful not to increase it twice:
statement
: someParserRule {loc++;}
| // ...
;
someParserRule
: SomeLexerRule someOtherParserRule {loc++;}
;
EDIT
I just noticed that in the title of your question you asked if this can be done during lexing. That won't be possible. Let's say a LoC would always end with a ';'. During lexing, you wouldn't be able to make a distinction between a ';' after, say, an assignment (which is a single LoC), and the 2 ';'s inside a for(int i = 0; i < n; i++) { ... } statement (which wouldn't be 2 LoC).
In the C target the data structure ANTLR3_INPUT_STREAM has a getLine() function which returns the current line from the input stream. It seems the Java version of this is CharStream.getLine(). You should be able to call this at any time and get the current line in the input stream.
Use a visitor to visit the CompilationUnit context, then context.stop.getLine() will give you the last line number of the compilation unit context.
#Override public Integer visitCompilationUnit(#NotNull JAVAParser.CompilationUnitContext ctx) {
return ctx.stop.getLine();
}
I am trying to write this code:
for (i = 0; i <= CONST - 1'b1; i = i + 1'b1)
begin : loop_inst
if (i < 3)
begin
if (changed[i] & !done_q[i])
begin
writedata[3-i] = en[i];
writedata[2-i:0] = readdata[2-i:0];
writedata[15:4-i] = readdata[15:4-i];
end
end
else
...
Basically, the location of the bit I am trying to write to (en) changes depending on which address I am talking to, depending on i. This code is not synthesizable because i is not a constant.
Is there any other workaround to this? The only workaround I know is writing out those three statements CONST times. I am hoping I DON'T have to do that in the end. Is there any other solution?
It looks like you're trying to copy readdata to writedata all the time, but fill in the LSBs with en if certain special case conditions are met. I'm also going to assume that the for loop you have is in an always block, and that you're intending to build combo logic.
The for loop as you've it written doesn't make much sense to me from a hardware perspective. A for loop is used for building arrays of logic, and as you've
written it you'll have at least 3 logic cones trying to set values on the entire writedata bus. (If it generates anything at all, it'll be some weird priority structure).
That said, it's probably the range selects that your compiler is complaining about, ie the writedata[2-i:0] rather than the writedata[3-i] = en[i]; (anything with : in the part select). If you want to do something along those lines, you can use 'indexed part selects' ( +: or -:) but there are better solutions in this case.
I'd rewrite it as follows - assuming I've assumed correctly :)
always #( /*whatever*/ ) begin
// default assignment
writedata = readdata;
// overwrite some bits in writedata for special cases
for(i=0; i<3; i++) begin
if( changed[i] & !done_q[i] )
writedata[3-i] = en[i];
end
end
In this code, I'm setting writedata to readdata, and then tweaking the resulting value of writedata if the special cases are in play. The for loop is building 3 logic cones, one for each of the bits in writedata[3:1]. I'd double-check if the bit mapping is what you intend -ie, mapping en[2:0] on to writedata[1:3].