How can I remove white spaces from a string in MIPS assembly language.?
371191
Without giving explicit syntax or writing it for you, here is the procedure it needs to follow:\nSet aside an equal or larger data area to copy the new string into.\nStep through the characters in the string checking for whitespace as you go.  \nIf it is whitespace, discard it else write it into the new string area.\nAfter all characters are tested, print the new string to the console.\n\nI hope that this will help you out.
976275
2x + 28 < with a line under it _ <_   6x\n\nsorry don't know how to add that special character in, but it is only one thing like the +\n between the 28 and 6
515796
go to: http://softwareclub.ws/v_split.html  i hope this will help a li'l bit
865202
Dim s as String\n\n'assuming s is not nothing\ns = IIF(s.length() > 30, s.substring(0, 30), s)
357099
void\nmain()\n{\n   if (printf("morgan&#92;n")) {\n   }\n}\n\n\nNo semi's -- no problem.
60802
"disp" or "display" functions\n"line" is helpful too\n\nTry to make a string x='*****' y='** **' \ndisp x,y \nETC....
1084791
You could use Instr() and Mid() to find the spaces and replace the character in the next position.\n\nBut the Split() function will eliminate the need to you search for spaces, and it will return each segment of the string in its own array element, so you can just UCase() the first character of each element, and then piece it back together.\n\nFunction ProperCase(asSourceString As String) As String\n    Dim pvSplit As Variant, psTemp As String, plCounter As Long\n    \n    pvSplit = Split(asSourceString, " ")\n    For plCounter = LBound(pvSplit) To UBound(pvSplit)\n        If Len(pvSplit(plCounter)) > 0 Then\n            pvSplit(plCounter) = UCase$(Left$(pvSplit(plCounter) , 1)) & Mid$(pvSplit(plCounter), 2)\n        End If\n        psTemp = psTemp & pvSplit(plCounter)\n        If plCounter < UBound(pvSplit) Then\n            psTemp = psTemp & " "\n        End If\n    Next\n    \n    ProperCase = psTemp\nEnd Function
780343
If you are in VBA in MS Access? Than you can cut the string and used the variable directly and then finish the string.\n\nstudTbl.Open "SELECT * FROM [studlst] where ([no]=" & i &")"
382045
Maybe you could make a method to remove certain characters from a string like string removeChars( string in, string chars )\n\nThe first param is the string you want to change and the second is the characters that you want to remove from the string. The method would use a for loop that loops through the chars string and in the for loop you would have a while loop that keeps on looping while the input string contains the character in the chars string at the index of the for loop. I don't think I said that very well, but the code might look like this:\n\nstring removeChars( string in, string chars );\n{\nfor ( int i = 0; i < chars.length(); i++ )\n{\nchar theChar = (chars.c_str())[ i ];\nint f = in.find_first_of( theChar );\nwhile ( f != -1 )\n{\nin.replace( f, 1, "" );\n}\n}\n\nreturn in;\n}\n\nThen, somewhere else in the program, you could call it like this: string cleanString = removeChars( "(in}[)]", "(){}[]" ); Then cleanString would equal "in".
92611
Use the following syntaxis:\n\nLike *[#]*
38471
Different systems use a different end of line character in the file formatting.  When a file is made and the program writes it out there are a few different end of line characters to choose from (depending on your operating system).  The program is writing it out with an end of line character that the conputer you are viewing it on reads as a ^M rather than an end of line.\n\nAre you viewing it on the same system as your program is writing it?  If so, then look in the program that you are making the page in and find the preference for end of line and change it.
430203
TWO EMPTY MATCH-BOXES CONNECTED WITH LONG STRING.
150391
It is a strings proccessing alg which uses a failure function to improve the run time rather than bm alg\nrun time:m+n\nm:the length of the pattern\nn: the length of the text\nwe use tis algorithm to find the pat string in the text string..\nmore details \nmail me at ziko@math.com\ni can provid a pascal source for it
526248
There are many answers to this:\n\n1) Take the string, make a reverse copy of it and compare the strings\n\n2) Pointer to the start index of the string and end index of the string and start comparing till you get strlen/2\n\n.... for more solutions and code you can search the web
403823
the first character in the disk directory is replaced with a zero that makes it appear to be deleted
389395
no. \n\nthose programs actually count blank spaces. \n\nso in between each of these words is a blank space. and each blank space actually has an ASNI or ASCII conversion. i forget which, like a blank space actually on a web site is  \n\nyou count those and you get the word count.
780385
Try it out - you need a semicolon at the end of the first print statement for it to compile.\nIn java,\nit prints out:\n0000000000\n0123456789\n0246802468\n0369258147\n0482604826\n0505050505\n0628406284\n0741852963\n0864208642\n0987654321