Write an expression that results in a string consisting of the third through tenth characters of the string?
91807
Easy! Now since you didn't specify a language, I'll give you a sample of 3 languages.\n\nHere it is in Java:\n...\nString s = "Some string here";\nString subString = s.substring(2,9);\n...\nIn Delphi:\n...\nvar s, subString:String;\nbegin\n...\ns = "Some string here";\nsubString := MidStr(s,3,10);\n\nIn VB: \ndim s, subString as String\ns = "Some string here"\nsubString = Mid(s,3,10)\n\nIf you need any more help, don't hesitate to e-mail me on nate AT tsamail.co.za.
918056
There is probably a better way, or at least a more streamline way, of doing this, but off the top of my head here is what I would do:\n\n1. Create a loop to run through all the characters in your string.\n2. Test each character to see if it falls between ASCII values 47 - 58 (i.e. 0-9). If it does then the character is a number.\n3. If the first test fails then you know that it isn't a number; therefore, you know that it must be a character, so now test to see if it is a character between 'a' and 'd' (i.e. ASCII 97-100).\n\nHere is some sample code (hopefully it can help):\npublic static void main(String[] args){\nString teststr = "d3";\nchar x;\n\nSystem.out.println("Test String:" + teststr);\n\nfor(int i = 0; i < teststr.length(); i++){\nx = teststr.charAt(i);\nif(x >= 47 && x <= 58){\n// Store it however\nSystem.out.println(Integer.parseInt(""+ x));\n}\nelse if(x >= 97 && x <= 100){\nif(x == 'a')\nSystem.out.println("a=" + 1);\nelse if(x == 'b')\nSystem.out.println("b=" + 2);\nelse if(x == 'c')\nSystem.out.println("c=" + 3);\nelse if(x == 'd')\nSystem.out.println("d=" + 4);\n}\n}\n}
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
816710
Create an outer for loop that will loop from 7 down to 0, with a step of 2.\n\nCreate one inner loops.\n\nThe inner loop should take the value of i, place it in another variable and count down to 0.\nThen print out the dollar signs for each iteration of this loop.
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
666462
ZOTTF (0 1 2 3 4)\n\nNEXT:\n\nFSSENT... (5 6 7 8 9 10...)
997808
Build an Update Query. . .  \n\nTo capture the characters to the left of the comma, you would use:  Left([Name Field],(InStr(1,[Name Field],",")-1)))\n\nFor the characters to the right of the comma: \nRight([Name Field],(Len([Name Field])-(InStr(1,[Name Field],",")+1)))\n\nI couldn't respond to your email as it said that it wasn't confirmed.  After you confirm your email, try emailing me again and I'll explain the above.
1099552
x' € S - {x}\nworks fine...\n( replace the € by the usual "is an element of" symbol )
411292
static_cast<int>("A") = 65\n\ngives you the ASCII value of the character\n\nWhole program:==============================================\n\nchar array[25]={'A','B',.....'Z'}\n\ncout << "Enter number: ";\ncin >> number\n\nif(number <= 10){\n  cout << number;\n}\nelse\n cout << array[number- 10];\n\nThat is the basic program you need to write it will work just dress it up to look pretty
600562
Okay once again..\nStringBuffer b = buf.replace(6,10,"orld");\n\nThis replaces offset 6 to 10 with world... So offset 6 in your case is W so it would be:\nHello orld\n\nNow That is correct.. Now b.delete(0, 1)\nThat deletes the 0 offset which is H! only so it would be\nello\n\nNow the total StringBuffer Contents is:\nello orld\n\n\nSee why? Remember buffers is like a memory location that stores a buffer, so treat it as global scope in that case
1066435
it is seven letters long.\n"a string"
677729
2/3 the number that is not a variable
49759
A comparable question: How long is a piece of string?
537104
Nine (9) - before 10 and after 11\nOne (1) - before 2 and after 0\nOne (1) - before 2 and after 0
184
Use the CONCATENATE function, with one of the arguments being ", "\n\nThe formula you end up with should look something like this:\n=CONCATENATE(D3,", ",D5)
361979
if the last # is in the tenths place put it over 10
315617
void StrReverse4(char *str)\n{\n    if(*str)\n    {\n        StrReverse4(str+1);\n        putchar(*str);\n    }\n}\n\nTry this code it might work