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Place String Literal Before Variable


Place String Literal Before Variable

By : Shaynee-honor Monroe
Date : November 20 2020, 11:01 PM
this will help Probably your environment defines _UNICODE for compilation process and MessageBox macro is expanded to MessageBoxW. Try to replace your code with this:
code :
MessageBoxA(NULL, input, "You wrote", MB_OK);


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how to place a character literal in a python string

how to place a character literal in a python string


By : Vijayakumar RM
Date : March 29 2020, 07:55 AM
will help you chr(1) should work, as will "\x01", as will "\001". (Note that chr(1) already returns a string, so you don't need to do str(chr(1)).) In your example it looks like you have both "\001" and chr(1), so that won't work unless you have two of the characters in a row in your data.
You say the regex "doesn't seem to catch it", but you don't give an example of your input data, so it's impossible to say why.
When comparing a variable to a literal, should one place the literal on the left or right of the equals '==' operator?

When comparing a variable to a literal, should one place the literal on the left or right of the equals '==' operator?


By : Sudesh Dhawade
Date : March 29 2020, 07:55 AM
To fix this issue
The reasoning being that if you accidentally type = instead of ==, then the code will fail at compile.
code :
cl /Wall Test.c
test.c(3) : warning C4706: assignment within conditional expression
if (foo == true)
if (foo = true)
if (foo)
if (!foo)
if (x == MethodReturningBool())
if (MethodReturningBool() == x)
What is it called when you pass a string literal to a variable, then assign data from that string to a new variable like

What is it called when you pass a string literal to a variable, then assign data from that string to a new variable like


By : Ferdinand Cirel Adit
Date : March 29 2020, 07:55 AM
I hope this helps . Formatting can be helpful. If we unroll the line that's giving the error, we see:
code :
(function() {
   var str = "Hello", 
       length = str.length, 
       var lengthtwo = str.length*2;
 })();
(function() {
   var str = "Hello", 
       length = str.length, 
       lengthtwo = str.length*2;
 })();
(function() {
   var str = "Hello", 
       length = str.length;
   var lengthtwo = str.length*2;
 })();
Python string literal throwing errors with place holders

Python string literal throwing errors with place holders


By : ishansethi1995
Date : March 29 2020, 07:55 AM
should help you out I have a list as : token_found = ["abcdxyz", "1234567"] , curly brackets {, } should be doubled to print them literally:
code :
token_found = ["abcdxyz", "1234567"]
user_creds_string = """
UserA_A:
    - name: 'Abcd'
      value: '{0}'
UserB_B:
    - name: 'Abcd'
      value: '{1}'
headers:
  - name: 'Abcd'
    value: 'Kota %(xyz)'
    inputs:
      apop: {{'type':'hex-64', 'required': True}}
  - name: 'X-Broke'
    value: '%(BId) %(blahId)'
    inputs:
      UkID: {{'type':'hex-16', 'required': True}}
      blahId: {{'type':'hex-64', 'required': True}}
apis:
    """.format(token_found[0],token_found[1])

print(user_creds_string)
UserA_A:
    - name: 'Abcd'
      value: 'abcdxyz'
UserB_B:
    - name: 'Abcd'
      value: '1234567'
headers:
  - name: 'Abcd'
    value: 'Kota %(xyz)'
    inputs:
      apop: {'type':'hex-64', 'required': True}
  - name: 'X-Broke'
    value: '%(BId) %(blahId)'
    inputs:
      UkID: {'type':'hex-16', 'required': True}
      blahId: {'type':'hex-64', 'required': True}
apis:
How to properly concatenate string literal to string variable and reassign to variable

How to properly concatenate string literal to string variable and reassign to variable


By : user2809996
Date : March 29 2020, 07:55 AM
this will help Firstly, I think you intended to do wordWithoutVowels += letter, not the entire userWord. Secondly, that expression is the same as wordWithoutVowels = wordWithoutVowels + userWord, which means that wordWithoutVowels needs to be defined before it.
Simply add the following before the for loop
code :
wordWithoutVowels = ''
if letter not in ['A','E','I','O','U']:
    wordWithoutVowels += letter 
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