If you really want to implement this as a macro then you need to remove the return from inside the macro. You can’t assign the “result” of your macro to a variable as it is not a single statement. #include <cstdio> #define IPV4_ZEROVAL_CHECK(ip, result) {int d1,d2,d3,d4; result = (4 == sscanf(ip, “%d.%d.%d.%d”, &d1, &d2, &d3, … Read more
You can’t use this macro outside of a function, because it’s an arbitrary expression, that’s why you’re getting an error. Just move the invocation of the macro into function scope and it will work: #define MAX(x, y, r) ((x) > (y) ? (r = x) : (r = y)) int x = 10; int y … Read more
Macros (#define MACRO …) are processed by the C preprocessor before the actuel compile process takes place. So the compiler only “sees” this once the file has been preprocessed: int main() { // your code goes here int k = 0; scanf(“%d”, &k); switch (k) { case 1: break; case 2: break; case 3: break; … Read more
Here’s a sample of how to do that: Sub createLink() Dim lastRow As Integer, sheetCount As Integer, myRange As Excel.Range, c As Excel.Range lastRow = Cells.Find(“*”, SearchOrder:=xlByRows, SearchDirection:=xlPrevious).Row sheetCount = Application.Sheets.Count Set myRange = Excel.ThisWorkbook.Sheets(“Sheet1”).Range(“B1:B” & lastRow) For Each c In myRange For x = 1 To sheetCount If Worksheets(x).Name = c.Value Then Excel.ThisWorkbook.Sheets(“Sheet1”).Hyperlinks.Add Anchor:=c, … Read more
The macro you defined lacks brackets to keep the arithmetic working as you want. Remember preprocessor macros are doing text replacement solely. So what you’ll get from calling it as you shown expands to int answer (2 + 4 * 2 + 4); and according operator precedence the result is 14. Write your macro as … Read more
The idea is “recursively defining the problem down to 2-bit values”: 00, 01, 10, 11. They’re not recursive macros, but does represent the technique of recursive decomposition of the problem. The arrangement of macros as a cascade attacks the problem of generating the table of 2^8 values by solving the problem for 2-bits (generating 4 … Read more
Because ‘<‘ is a binary operator and it cannot be evaluated without an operand on either side. You can verify this without macros simply by attempting to compile the following code: bool LessThan( int a, int b ) { return( a (<) b ); } At the very least you should see “expected an expression” … Read more
If you stick to the Windows platform you can do this with Auto-it. It is a scripting language which allows you to move the mouse pointer, read pixel colours etc. I used it to cheat at Evony years ago, before everybody else also started cheating with YAEB and I lost (interest). solved Theoretically speaking, is … Read more
template<typename T> T xtime(T x) { return ((x<<1) ^ (((x>>7) & 1) * 0x1b)); } solved Converting a macro into a function
Here is the macro: #define IS_ALNUM(x) (((x)>=’a’ && (x) <= ‘z’)) ||((x)>=’A’ && (x) <= ‘Z’)) || (((x)>=’0′ && (x) <= ‘9’))) It tests if it is Between a and z Between A and Z Between 0 and 9 Quite simple 2 solved find the input is alphanumeric or not with using #define macro preprocessor … Read more
Macros are just replaced by their definition. When you #define k 1024; and write if(k==1024)… what the compiler actually sees is: if(1024; == 1024) … which doesn’t compile. The compiler doesn’t complain because sometimes you may actually want to add a semicolon (; is called semicolon, not colon, which is 🙂 to your macro. solved … Read more
How to create a combination generator where order does not matter but being limited from a specific range of Sum? Using Excel VBA macro [closed] solved How to create a combination generator where order does not matter but being limited from a specific range of Sum? Using Excel VBA macro [closed]
Here, this will do what you want. Option Explicit Sub FileListingAllFolder() Dim pPath As String Dim FlNm As Variant Dim ListFNm As New Collection ‘ create a collection of filenames Dim OWb As Workbook Dim ShtCnt As Integer Dim Sht As Integer Dim MWb As Workbook Dim MWs As Worksheet Dim i As Integer ‘ … Read more
In the macro #define prod(a,b) ((a>b)?a*a:b*b) a and b are referenced more than once. So prod(p1++,q1++) becomes (p1++ > q1++) ? p1++*p1++ : q1++*q2++ Note that this is very different than calling the function prod1(p1++, q1++) which only increments p1 and q1 once. And does so predictably. 2 solved Different output from macros and function … Read more
From the standard N4687: 20.5.4.3.2 Macro names 1 A translation unit that includes a standard library header shall not #define or #undef names declared in any standard library header. 2 A translation unit shall not #define or #undef names lexically identical to keywords, to the identifiers listed in Table 4, or to the attribute-tokens described … Read more