sizeof

I happened to stumble across this piece of code. int x(int a){ std::cout<<a<<std::endl; return a + 1; } int main() { std::cout<<sizeof(x(20))<<std::endl; return 0; } I expected it to print 20 followed 4. But it just prints 4. Why does it happen so? Isn't it incorrect to optimize out a function, that has a side effect (printing to IO/file etc)? sizeof is a compile-time operator, and the operand is never evaluated. sizeof is actually an operator and it is evaluated in compile-time. The compiler can evaluate it because the size of the return type of x is fixed; it cannot change during program

// PEOperate.cpp: implementation of the PEOperate class. // ////////////////////////////////////////////////////////////////////// #include "PEOperate.h" ////////////////////////////////////////////////////////////////////// // Construction/Destruction ////////////////////////////////////////////////////////////////////// #include "windows.h" #include "stdio.h" #include "string.h" #define MESSAGEBOXADDR 0x77D507EA #define SHELLCODELENGTH 0x12 BYTE shellCode[]={ 0x6A,00,0x6A,00,0x6A,00,0x6A,00, 0xE8,00,00,00,00, 0xE9,00,00,00,00 }; //加载PE文件到内存中 LPVOID ReadPEFile(LPSTR lpszFile) { FILE *pFile =

Suppose I have a class A that does not inherit from anything, has no virtual methods, and has exactly one variable of type T. Does C++ guarantee sizeof(A) == sizeof(T) ? EDIT: Also if T were a composite type, would it make a difference? No, it might be more than sizeof(T) due to padding. Seth Carnegie I don't think it explicitly guarantees it, but I don't think it would ever be different. I think C++ should guarantee sizeof(A) == sizeof(T). Consider bellow situation, C++ should make it works just like in C: A a[10]; T t[10]; A * ap = (A *) t; T * tp = (T *) a; memcpy(ap, tp, sizeof(*ap)); 来源：

char a, b; printf("%d", sizeof(a+b)); What will printf write to the screen? I thought because sizeof(char)=1, that sizeof(a+b) will be also 1, but it turned out to be 4. I don't understand this, why does it write 4 if we are adding two chars? In C language operands of almost all arithmetic operators are subjected to implicit conversions called usual arithmetic conversions or, in this case, integer promotions . Operands of type char are promoted to type int and the actual addition is performed within the domain of int (or unsigned int , depending on the properties of char on that platform). So

Are both these codes the same char ch = 'a'; printf("%d", ch); Will it print a garbage value? I am confused about this printf("%d", '\0'); Will this print 0 or garbage value? Because when i do this printf("%d", sizeof('\n')); It prints 4. Why is sizeof('\n') 4 bytes? The same thing in C++ prints 1 bytes. Why is that? So here's the main question in c language is printf("%d", '\0') supposed to print 0 and in C++ printf("%d", '\0') supposed to print garbage? %d prints an integer: it will print the ascii representation of your character. What you need is %c : printf("%c", ch); printf("%d", '\0');