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This question already has an answer here: Cannot implicitly convert type 'int' to 'short' [duplicate] 9 answers Following expression is ok short d = ("obj" == "obj" ) ? 1 : 2; But when you use it like below, syntax error occurs short d = (DateTime.Now == DateTime.Now) ? 1 : 2; Cannot implicitly convert type 'int' to 'short'. An explicit conversion exists (are you missing a cast?) Can anyone explain why this is so? Is there a difference between comparing string-to-string and datetime-to-datetime in a ternary operator, why? I would be grateful if you could help me. C# language specification,

int i=-3, j=2, k=0, m; m = ++i || ++j && ++k; printf("%d, %d, %d, %d\n", i, j, k, m); Since ++ has more precedence than || and && in C, they are evaluated first and therefore the expression becomes m = -2 || 3 && 1 . Now you can apply short circuiting but that produces incorrect answer. Why is that? Precedence ≠ order of evaluation. The short-circuiting behavior of || and && means that their left-hand sides are evaluated first, and If the LHS of || evaluates to true (nonzero), the RHS is not evaluated (because the expression will be true no matter what the RHS is) If the LHS of && evaluates to

When I run this very simple recursive code, I found out that when the function "recursCheck()" is using integer parameters my memory jumps to 200 MB in Win32 mode and 45 MB in x64 mode, and when the parameter is of type "short" the used memory is 4.7 MB Win32 mode and 1.7 MB for x64. sizeof(int) shows 4 bytes and sizeof(short) shows 2 bytes. How is this possible - 1 000 000 x 2 bytes = 2 MB and 1 000 000 x 4 bytes = 4 MB NOT 200 MB !!! So why this problem appers(short vs int) and why Win32 mode takes less memory than x64? I`m using Visual Studio 2013, 64 bit OS Windows7, CPU i5, 4 GB RAM. The

I want to know what is the reason of the output of the following codes: unsigned short a=10,aa=-1; if(a>-1) printf("surprise"); else printf(" No surprise"); This gives output "Surprise" unsigned int a=10,aa=-1; if(a>-1) printf("surprise"); else printf("No surprise"); This gives output "No Surprise" and unsigned short a=10,aa=-1; if(a>aa) printf("surprise"); else printf("No surprise"); This gives the output "No Surprise" willus See this Stack Exchange question: In a C expression where unsigned int and signed int are present, which type will be promoted to what type? In the response from

I'm trying to convert a HEX number into a short (2 Bytes) using C++ everything is OK except for one thing... signed conversion from short to Byte (last test) i found this question and couldn't really benefit from it: portable signed/unsigned byte cast,C++ here are my tests: // test 1 - positive B2Short (success) byte *b = new byte[2]; b[0] = 0x10; //low byte b[1] = 0x00; //heigh byte signed short test = 0; test = ByteToShort(b); cout << test << endl; // test 2 - negative B2Short (success) b[0] = 0xF0; //low byte b[1] = 0xFF; //heigh byte test = 0; test = ByteToShort(b); cout << test << endl; /

I understand that the 'natural size' is the width of integer that is processed most efficiently by a particular hardware. When using short in an array or in arithmetic operations, the short integer must first be converted into int . Q: What exactly determines this 'natural size'? I am not looking for simple answers such as If it has a 32-bit architecture, it's natural size is 32-bit I want to understand why this is most efficient, and why a short must be converted before doing arithmetic operations on it. Bonus Q: What happens when arithmetic operations are conducted on a long integer? the

I'm trying to write a batch file in Windows XP that takes in a fully-qualified path name and outputs the 8.3 short name version... @echo off echo "%~s1" I have come across one particular case where this outputs an incorrect path and file... C:\>test.bat "C:\Documents and Settings\angus\Local Settings\Temporary Internet Files\Content.IE5\2JSTM34V\62[1].ja2" "C:\DOCUME~1\angus\LOCALS~1\TEMPOR~1\Content.IE5\2JSTM34V\62_1_~1.JA2M34V\62[1].ja2" Note that the above output ("C:\DOCUME~1\angus\LOCALS~1\TEMPOR~1\Content.IE5\2JSTM34V\62_1_~1.JA2M34V\62[1].ja2") does not exist. If I remove the ".JA2M34V