expansion

Variable not recognized in else statement in a bat file

＊爱你＆永不变心＊ — Thu, 11 Feb 2021 09:14:49 +0000

Variable not recognized in else statement in a bat file ＊爱你＆永不变心＊ 2021-02-11 17:14:49

问题

I have the follow code:

ECHO OFF
:LOOP
set c=
setlocal ENABLEDELAYEDEXPANSION
tasklist | find /i "calc.exe" >nul 2>&1
IF %ERRORLEVEL% == 0 (
  ECHO easymeetingOnCall is running
  taskkill /IM calc.exe /F
  set c=true
  Timeout /T 10
  goto LOOP
) ELSE (
  IF "!c!" == "true" (
    ECHO l'applicazione easymeetingOnCall è spenta.
    Timeout /T 10
    goto exitLoop
  )
  ECHO easymeetingOnCall is not running
  Timeout /T 5 /Nobreak
  GOTO LOOP
)
:exitLoop

My problem is that the follow condition

IF "!c!" == "true"

Into the else statement, is not recognized. If I write echo !c! It doesn't output any result. I set enabledelayedexpansion before, so I don't know why of this behavior.

Could you help me? Thanks to all.

回答1:

Look carefully at the following extraction from your code.

ECHO OFF
:LOOP
set c=
....
set c=true
goto :LOOP

You are explicitly telling the if statement to goto :loop where directly after you set c= which now gives it no value. Move the set c= to above to label to retain the value you've set.

I however suggest a few changes. you can get away without delayedexpansion and do actually require all the for loops. Using taskkill to explicitly search for the app is better than listing all, also already mentioned to you in a comment by @Mofi. Lastly, it is not really good practice to set single character variables, though it does not always cause issues, I would suggest using multple character variable names. I just changed your variable %c% to %_c%

@echo off & set _c=
:loop
tasklist /FI "IMAGENAME eq Calc.exe" | find /i "Calc"
goto :_%errorlevel%
:_0
  ECHO easymeetingOnCall is running
  taskkill /IM calc.exe /F
  set _c=true
  Timeout /T 10
  goto loop
:_1
  IF "%_c%" == "true" (
    ECHO l'applicazione easymeetingOnCall è spenta.
    Timeout /T 10
    goto exitloop
  )
  ECHO easymeetingOnCall is not running
  Timeout /T 5 /Nobreak
  goto loop
  :exitloop

Edit, as you wanted to do this with a list:

@echo off & set _c=
set "list=calc.exe StickyNot.exe wordpad.exe"
for %%a in (%list%) do call :loop %%a
goto :eof
:loop
set task=%1
 tasklist /FI "IMAGENAME eq %1" | find /i "%1"
   if not errorlevel 1 (
   ECHO easymeetingOnCall is running %1
   taskkill /IM %1 /F
   set _c=true
   Timeout /T 10
)
if "%_c%" == "true" (
    ECHO l'applicazione easymeetingOnCall è spenta. %1
    Timeout /T 10
    goto exitloop
  )
    echo easymeetingOnCall is not running %1
    Timeout /T 5 /Nobreak
    goto :loop
  :exitloop

Be aware, if the first process is not running, it will loop forever until it is found, then only kill it and go to the next process. If you do not want that, then change goto :loop in the last line to exit loop.

来源：https://stackoverflow.com/questions/63357127/variable-not-recognized-in-else-statement-in-a-bat-file

标签

windows

batch-file

cmd

scope

expansion

Variable not recognized in else statement in a bat file

一笑奈何 — Thu, 11 Feb 2021 09:11:32 +0000

Variable not recognized in else statement in a bat file 一笑奈何 2021-02-11 17:11:32

问题

I have the follow code:

ECHO OFF
:LOOP
set c=
setlocal ENABLEDELAYEDEXPANSION
tasklist | find /i "calc.exe" >nul 2>&1
IF %ERRORLEVEL% == 0 (
  ECHO easymeetingOnCall is running
  taskkill /IM calc.exe /F
  set c=true
  Timeout /T 10
  goto LOOP
) ELSE (
  IF "!c!" == "true" (
    ECHO l'applicazione easymeetingOnCall è spenta.
    Timeout /T 10
    goto exitLoop
  )
  ECHO easymeetingOnCall is not running
  Timeout /T 5 /Nobreak
  GOTO LOOP
)
:exitLoop

My problem is that the follow condition

IF "!c!" == "true"

Into the else statement, is not recognized. If I write echo !c! It doesn't output any result. I set enabledelayedexpansion before, so I don't know why of this behavior.

Could you help me? Thanks to all.

回答1:

Look carefully at the following extraction from your code.

ECHO OFF
:LOOP
set c=
....
set c=true
goto :LOOP

You are explicitly telling the if statement to goto :loop where directly after you set c= which now gives it no value. Move the set c= to above to label to retain the value you've set.

@echo off & set _c=
:loop
tasklist /FI "IMAGENAME eq Calc.exe" | find /i "Calc"
goto :_%errorlevel%
:_0
  ECHO easymeetingOnCall is running
  taskkill /IM calc.exe /F
  set _c=true
  Timeout /T 10
  goto loop
:_1
  IF "%_c%" == "true" (
    ECHO l'applicazione easymeetingOnCall è spenta.
    Timeout /T 10
    goto exitloop
  )
  ECHO easymeetingOnCall is not running
  Timeout /T 5 /Nobreak
  goto loop
  :exitloop

Edit, as you wanted to do this with a list:

@echo off & set _c=
set "list=calc.exe StickyNot.exe wordpad.exe"
for %%a in (%list%) do call :loop %%a
goto :eof
:loop
set task=%1
 tasklist /FI "IMAGENAME eq %1" | find /i "%1"
   if not errorlevel 1 (
   ECHO easymeetingOnCall is running %1
   taskkill /IM %1 /F
   set _c=true
   Timeout /T 10
)
if "%_c%" == "true" (
    ECHO l'applicazione easymeetingOnCall è spenta. %1
    Timeout /T 10
    goto exitloop
  )
    echo easymeetingOnCall is not running %1
    Timeout /T 5 /Nobreak
    goto :loop
  :exitloop

来源：https://stackoverflow.com/questions/63357127/variable-not-recognized-in-else-statement-in-a-bat-file

标签

Do we need to call wordfree upon wordexp failure?

对着背影说爱祢 — Mon, 08 Feb 2021 11:38:27 +0000

Do we need to call wordfree upon wordexp failure? 对着背影说爱祢 2021-02-08 19:38:27

问题

Do we need to call wordfree upon wordexp failure? Calling wordfree seems to segfault in some cases (eg when wordfree returns error code with string is "foo 'bar"). This isn't clear from man page, and I've seen wordfree used in some error cases.

回答1:

According to the GNU's manual example, it should be called on error only if WRDE_NOSPACE was returned:

switch (wordexp (program, &result, 0))
{
case 0:         /* Successful.  */
  break;
case WRDE_NOSPACE:
  /* If the error was WRDE_NOSPACE,
     then perhaps part of the result was allocated.  */
  wordfree (&result);
default:                    /* Some other error.  */
  return -1;
}

来源：https://stackoverflow.com/questions/23120084/do-we-need-to-call-wordfree-upon-wordexp-failure

标签

posix

expansion

Macro stepper in DrRacket

拟墨画扇 — Wed, 27 Jan 2021 00:00:49 +0000

Macro stepper in DrRacket 拟墨画扇 2021-01-27 08:00:49

问题

On the link http://www.ccs.neu.edu/home/ryanc/macro-stepper/tutorial.html there are instructions for working with the macro stepper.

However, when I'm going to try it, I can't get the second expansion of myor in the definition of nonzero? function, only the first. Also, I have no buttons "Previous term" and "Next term".

So my question is: how i must configure macro stepper to get the second expansion, like in tutorial?

回答1:

I'm assuming that your source program looked something like this:

#lang racket
(define-syntax myor
  (syntax-rules ()
    [(myor e) e]
    [(myor e1 . es)
     (let ([r e1]) (if r r (myor . es)))]))
(define (nonzero? r)
  (myor (negative? r)
        (positive? r)))

Short story: use syntax-case rather than syntax-rules for the moment; there appears to be some bug associated with the Macro stepper and syntax-rules. I've sent along a bug report to the Racket developers, so hopefully this will be fixed soon. The syntax-case version of the above program looks like this.

#lang racket

(define-syntax (myor stx)
  (syntax-case stx ()
    [(_ e) #'e]
    [(_ e1 . es)
     #'(let ([r e1]) (if r r (myor . es)))]))

(define (nonzero? r)
  (myor (negative? r)
        (positive? r)))

Below is the longer story...

When I run your program under the pre-release of 5.2.1, I see the following in the Macro Stepper, with Macro hiding set to "Standard":

(module anonymous-module racket
  (#%module-begin
   (define-syntax myor
     (syntax-rules () [(myor e) e] [(myor e1 . es) (let ([r e1]) (if r r (myor . es)))]))
   (define (nonzero? r)
     (let:26 ([r:26 (negative? r)]) (if:26 r:26 r:26 (myor:26 (positive? r)))))))

which looks wrong. It expands only one use of myor out to uses of if. Very odd!

Let's see what things look like under Racket 5.2...

(module anonymous-module racket
  (#%module-begin
   (define-syntax myor
     (syntax-rules () [(myor e) e] [(myor e1 . es) (let ([r e1]) (if r r (myor . es)))]))
   (define (nonzero? r) (let ([r (negative? r)]) (if r r (myor (positive? r)))))))

Ah. Ok, I can confirm that I see the same problem that you see in Racket 5.2, as well as the pre-release.

The bug seems related to the behavior of the "Macro hiding" feature, which tries to not overwhelm you with the full expansion when it is set to Standard. If you set it to "Disabled", you'll see that the macro debugger will show the expansion in its full, unvarnished glory, and it does include the expansion that we expect to see:

(module anonymous-module racket
  (#%module-begin
   (define-syntaxes (myor)
     (lambda (x)
        ; ... I'm omitting the content here: it's way too long.
     ))
   (define-values:20 (nonzero?)
     (lambda:21 (r) (let-values:22 (((r) (#%app:23 negative? r))) (if r r (#%app:24 positive? r)))))))

I'll write a bug report and send it to the Racket developers.

If you write your macro using syntax-case, as opposed to syntax-rules, it appears to work better with the Macro Stepper.

#lang racket

(define-syntax (myor stx)
  (syntax-case stx ()
    [(_ e) #'e]
    [(_ e1 . es)
     #'(let ([r e1]) (if r r (myor . es)))]))

(define (nonzero? r)
  (myor (negative? r)
        (positive? r)))

When I step through this, it appears to work a lot better. So whatever is triggering the bug, it appears to be some interaction with the Macro Stepper and syntax-rules. So try using syntax-case instead.

回答2:

It worked for me without having to do anything additional. Try with the newest version of Racket, also try choosing a different language from the menu, and make sure that debugging is selected for the language you select.

来源：https://stackoverflow.com/questions/8961762/macro-stepper-in-drracket

标签

Macro stepper in DrRacket

一笑奈何 — Tue, 26 Jan 2021 23:54:50 +0000

Macro stepper in DrRacket 一笑奈何 2021-01-27 07:54:50

问题

On the link http://www.ccs.neu.edu/home/ryanc/macro-stepper/tutorial.html there are instructions for working with the macro stepper.

However, when I'm going to try it, I can't get the second expansion of myor in the definition of nonzero? function, only the first. Also, I have no buttons "Previous term" and "Next term".

So my question is: how i must configure macro stepper to get the second expansion, like in tutorial?

回答1:

I'm assuming that your source program looked something like this:

#lang racket
(define-syntax myor
  (syntax-rules ()
    [(myor e) e]
    [(myor e1 . es)
     (let ([r e1]) (if r r (myor . es)))]))
(define (nonzero? r)
  (myor (negative? r)
        (positive? r)))

#lang racket

(define-syntax (myor stx)
  (syntax-case stx ()
    [(_ e) #'e]
    [(_ e1 . es)
     #'(let ([r e1]) (if r r (myor . es)))]))

(define (nonzero? r)
  (myor (negative? r)
        (positive? r)))

Below is the longer story...

When I run your program under the pre-release of 5.2.1, I see the following in the Macro Stepper, with Macro hiding set to "Standard":

(module anonymous-module racket
  (#%module-begin
   (define-syntax myor
     (syntax-rules () [(myor e) e] [(myor e1 . es) (let ([r e1]) (if r r (myor . es)))]))
   (define (nonzero? r)
     (let:26 ([r:26 (negative? r)]) (if:26 r:26 r:26 (myor:26 (positive? r)))))))

which looks wrong. It expands only one use of myor out to uses of if. Very odd!

Let's see what things look like under Racket 5.2...

(module anonymous-module racket
  (#%module-begin
   (define-syntax myor
     (syntax-rules () [(myor e) e] [(myor e1 . es) (let ([r e1]) (if r r (myor . es)))]))
   (define (nonzero? r) (let ([r (negative? r)]) (if r r (myor (positive? r)))))))

Ah. Ok, I can confirm that I see the same problem that you see in Racket 5.2, as well as the pre-release.

(module anonymous-module racket
  (#%module-begin
   (define-syntaxes (myor)
     (lambda (x)
        ; ... I'm omitting the content here: it's way too long.
     ))
   (define-values:20 (nonzero?)
     (lambda:21 (r) (let-values:22 (((r) (#%app:23 negative? r))) (if r r (#%app:24 positive? r)))))))

I'll write a bug report and send it to the Racket developers.

If you write your macro using syntax-case, as opposed to syntax-rules, it appears to work better with the Macro Stepper.

#lang racket

(define-syntax (myor stx)
  (syntax-case stx ()
    [(_ e) #'e]
    [(_ e1 . es)
     #'(let ([r e1]) (if r r (myor . es)))]))

(define (nonzero? r)
  (myor (negative? r)
        (positive? r)))

回答2:

来源：https://stackoverflow.com/questions/8961762/macro-stepper-in-drracket

标签

Macro stepper in DrRacket

倖福魔咒の — Tue, 26 Jan 2021 23:52:50 +0000

Macro stepper in DrRacket 倖福魔咒の 2021-01-27 07:52:50

问题

On the link http://www.ccs.neu.edu/home/ryanc/macro-stepper/tutorial.html there are instructions for working with the macro stepper.

However, when I'm going to try it, I can't get the second expansion of myor in the definition of nonzero? function, only the first. Also, I have no buttons "Previous term" and "Next term".

So my question is: how i must configure macro stepper to get the second expansion, like in tutorial?

回答1:

I'm assuming that your source program looked something like this:

#lang racket
(define-syntax myor
  (syntax-rules ()
    [(myor e) e]
    [(myor e1 . es)
     (let ([r e1]) (if r r (myor . es)))]))
(define (nonzero? r)
  (myor (negative? r)
        (positive? r)))

#lang racket

(define-syntax (myor stx)
  (syntax-case stx ()
    [(_ e) #'e]
    [(_ e1 . es)
     #'(let ([r e1]) (if r r (myor . es)))]))

(define (nonzero? r)
  (myor (negative? r)
        (positive? r)))

Below is the longer story...

When I run your program under the pre-release of 5.2.1, I see the following in the Macro Stepper, with Macro hiding set to "Standard":

(module anonymous-module racket
  (#%module-begin
   (define-syntax myor
     (syntax-rules () [(myor e) e] [(myor e1 . es) (let ([r e1]) (if r r (myor . es)))]))
   (define (nonzero? r)
     (let:26 ([r:26 (negative? r)]) (if:26 r:26 r:26 (myor:26 (positive? r)))))))

which looks wrong. It expands only one use of myor out to uses of if. Very odd!

Let's see what things look like under Racket 5.2...

(module anonymous-module racket
  (#%module-begin
   (define-syntax myor
     (syntax-rules () [(myor e) e] [(myor e1 . es) (let ([r e1]) (if r r (myor . es)))]))
   (define (nonzero? r) (let ([r (negative? r)]) (if r r (myor (positive? r)))))))

Ah. Ok, I can confirm that I see the same problem that you see in Racket 5.2, as well as the pre-release.

(module anonymous-module racket
  (#%module-begin
   (define-syntaxes (myor)
     (lambda (x)
        ; ... I'm omitting the content here: it's way too long.
     ))
   (define-values:20 (nonzero?)
     (lambda:21 (r) (let-values:22 (((r) (#%app:23 negative? r))) (if r r (#%app:24 positive? r)))))))

I'll write a bug report and send it to the Racket developers.

If you write your macro using syntax-case, as opposed to syntax-rules, it appears to work better with the Macro Stepper.

#lang racket

(define-syntax (myor stx)
  (syntax-case stx ()
    [(_ e) #'e]
    [(_ e1 . es)
     #'(let ([r e1]) (if r r (myor . es)))]))

(define (nonzero? r)
  (myor (negative? r)
        (positive? r)))

回答2:

来源：https://stackoverflow.com/questions/8961762/macro-stepper-in-drracket

标签

Replacing 'source file' with its content, and expanding variables, in bash

﹥＞﹥吖頭↗ — Mon, 07 Dec 2020 23:55:53 +0000

Replacing 'source file' with its content, and expanding variables, in bash ﹥＞﹥吖頭↗ 2020-12-08 07:55:53

问题

In a script.sh,

source a.sh
source b.sh

CMD1
CMD2
CMD3

how can I replace the source *.sh with their content (without executing the commands)? I would like to see what the bash interpreter executes after sourcing the files and expanding all variables.

I know I can use set -n -v or run bash -n -v script.sh 2>output.sh, but that would not replace the source commands (and even less if a.sh or b.sh contain variables).

I thought of using a subshell, but that still doesn't expand the source lines. I tried a combination of set +n +v and set -n -v before and after the source lines, but that still does not work.

I'm going to send that output to a remote machine using ssh. I could use <<output.sh to pipe the content into the ssh command, but I can't log as root onto the remote machine, but I am however a sudoer. Therefore, I thought I could create the script and send it as a base64-encoded string (using that clever trick ) base64 script | ssh remotehost 'base64 -d | sudo bash'

Is there a solution? Or do you have a better idea?

回答1:

You can do something like this:

inline.sh:

#!/usr/bin/env bash
while read line; do
    if [[ "$line" =~ (\.|source)\s+.+ ]]; then
        file="$(echo $line | cut -d' ' -f2)"
        echo "$(cat $file)"
    else
      echo "$line"
    fi
done < "$1"

Note this assumes the sourced files exist, and doesn't handle errors. You should also handle possible hashbangs. If the sourced files contain themselves source, you need to apply the script recursively, e.g. something like (not tested):

while egrep -q '^(source|\.)' main.sh; do
    bash inline.sh main.sh > main.sh
done

Let's test it

main.sh:

source a.sh
. b.sh

echo cc
echo "$var_a $var_b"

a.sh:

echo aa
var_a="stack"

b.sh:

echo bb
var_b="overflow"

Result:

bash inline.sh main.sh

echo aa
var_a="stack"
echo bb
var_b="overflow"

echo cc
echo "$var_a $var_b"

bash inline.sh main.sh | bash

aa
bb
cc
stack overflow

BTW, if you just want to see what bash executes, you can run

bash -x [script]

or remotely

ssh user@host -t "bash -x [script]"

来源：https://stackoverflow.com/questions/37531927/replacing-source-file-with-its-content-and-expanding-variables-in-bash

标签

Replacing 'source file' with its content, and expanding variables, in bash

点点圈 — Mon, 07 Dec 2020 23:55:09 +0000

Replacing 'source file' with its content, and expanding variables, in bash 点点圈 2020-12-08 07:55:09

问题

In a script.sh,

source a.sh
source b.sh

CMD1
CMD2
CMD3

how can I replace the source *.sh with their content (without executing the commands)? I would like to see what the bash interpreter executes after sourcing the files and expanding all variables.

I know I can use set -n -v or run bash -n -v script.sh 2>output.sh, but that would not replace the source commands (and even less if a.sh or b.sh contain variables).

I thought of using a subshell, but that still doesn't expand the source lines. I tried a combination of set +n +v and set -n -v before and after the source lines, but that still does not work.

Is there a solution? Or do you have a better idea?

回答1:

You can do something like this:

inline.sh:

#!/usr/bin/env bash
while read line; do
    if [[ "$line" =~ (\.|source)\s+.+ ]]; then
        file="$(echo $line | cut -d' ' -f2)"
        echo "$(cat $file)"
    else
      echo "$line"
    fi
done < "$1"

while egrep -q '^(source|\.)' main.sh; do
    bash inline.sh main.sh > main.sh
done

Let's test it

main.sh:

source a.sh
. b.sh

echo cc
echo "$var_a $var_b"

a.sh:

echo aa
var_a="stack"

b.sh:

echo bb
var_b="overflow"

Result:

bash inline.sh main.sh

echo aa
var_a="stack"
echo bb
var_b="overflow"

echo cc
echo "$var_a $var_b"

bash inline.sh main.sh | bash

aa
bb
cc
stack overflow

BTW, if you just want to see what bash executes, you can run

bash -x [script]

or remotely

ssh user@host -t "bash -x [script]"

来源：https://stackoverflow.com/questions/37531927/replacing-source-file-with-its-content-and-expanding-variables-in-bash

标签

Approximation of e^x using Maclaurin Series in Python

拈花ヽ惹草 — Mon, 01 Jun 2020 04:40:56 +0000

Approximation of e^x using Maclaurin Series in Python 拈花ヽ惹草 2020-06-01 12:40:56

问题

I'm trying to approximate e^x using the Maclaurin series in a function called my_exp(x), I believe everything I've done so far is right but I'm getting incorrect approximations for whatever number I try.

import math
for i in range (x):
    exp = 1 + ((x**i)/math.factorial(i))
print(exp)

For example, whenever I try my_exp(12) I get 18614.926233766233 instead of 162754.79141900392 Help?

回答1:

Your problem is that the e^x series is an infinite series, and so it makes no sense to only sum the first x terms of the series.

def myexp(x):
  e=0
  for i in range(0,100): #Sum the first 100 terms of the series
    e=e+(x**i)/math.factorial(i)
  return e

You can also define the precision of your result and get a better solution.

def myexp(x):
    e=0
    pres=0.0001
    s=1
    i=1
    while s>pres:
        e=e+s
        s=(x**i)/math.factorial(i)
        i=i+1
    return e

回答2:

In order to accumulate the terms of the series, you need to replace the assignment to exp with a line such as:

exp = exp + ((x**i)/math.factorial(i))

来源：https://stackoverflow.com/questions/24915194/approximation-of-ex-using-maclaurin-series-in-python

标签

Approximation of e^x using Maclaurin Series in Python

天涯浪子 — Mon, 01 Jun 2020 04:40:11 +0000

Approximation of e^x using Maclaurin Series in Python 天涯浪子 2020-06-01 12:40:11

问题

import math
for i in range (x):
    exp = 1 + ((x**i)/math.factorial(i))
print(exp)

For example, whenever I try my_exp(12) I get 18614.926233766233 instead of 162754.79141900392 Help?

回答1:

Your problem is that the e^x series is an infinite series, and so it makes no sense to only sum the first x terms of the series.

def myexp(x):
  e=0
  for i in range(0,100): #Sum the first 100 terms of the series
    e=e+(x**i)/math.factorial(i)
  return e

You can also define the precision of your result and get a better solution.

def myexp(x):
    e=0
    pres=0.0001
    s=1
    i=1
    while s>pres:
        e=e+s
        s=(x**i)/math.factorial(i)
        i=i+1
    return e

回答2:

In order to accumulate the terms of the series, you need to replace the assignment to exp with a line such as:

exp = exp + ((x**i)/math.factorial(i))

来源：https://stackoverflow.com/questions/24915194/approximation-of-ex-using-maclaurin-series-in-python

标签