c-preprocessor

问题 Why is the #if condition in the following code fulfilled: #include <iostream> #define VALUE foo int main() { #if VALUE == bar std::cout << "WORKS!" << std::endl; #endif // VALUE } 回答1: The page on cppreference.com states: After all macro expansion and evaluation of defined and __has_include (since C++17) expressions, any identifier which is not a boolean literal is replaced with the number ​0​ (this includes identifiers that are lexically keywords, but not alternative tokens like and). So

问题 Let's say I have 2 sets of values for P_A, P_B, P_C as below #define X_P_A 2 #define X_P_B 3 #define X_P_C 4 #define Y_P_A 5 #define Y_P_B 6 #define Y_P_C 7 There are 3 types of users:- once that only need X variants, once that only need Y variants and once those may need both. eg #ifdef X #define P_A X_P_A #define P_B X_P_B #define P_C X_P_C #endif #ifdef Y #define P_A Y_P_A #define P_B Y_P_B #define P_C Y_P_C #endif Users that need both will make the decision at run time and call X_P_<> or

问题 Let's say I have 2 sets of values for P_A, P_B, P_C as below #define X_P_A 2 #define X_P_B 3 #define X_P_C 4 #define Y_P_A 5 #define Y_P_B 6 #define Y_P_C 7 There are 3 types of users:- once that only need X variants, once that only need Y variants and once those may need both. eg #ifdef X #define P_A X_P_A #define P_B X_P_B #define P_C X_P_C #endif #ifdef Y #define P_A Y_P_A #define P_B Y_P_B #define P_C Y_P_C #endif Users that need both will make the decision at run time and call X_P_<> or

问题 This question already has answers here : Should preprocessor instructions be on the beginning of a line? (5 answers) Closed 5 years ago . I have programmed C for quite a while now. During this time I have learned that it is a common convention to put the "#"-character that comes before preprocessor-directives at column one. Example: #include <stdio.h> int main(void) { #ifdef MACRO1 #ifdef MACRO2 puts("defined(MACRO1) && defined(MACRO2)"); #else puts("defined(MACRO1)"); #endif #else puts("

问题 I have a preprocessor macro defined in build settings FOO=BAR That value I want to massage into an Objective-C string literal that can be passed to a method. The following #define does not work, but it should demonstrate what I am trying to achieve: #define FOOLITERAL @"FOO" //want FOOLITERAL to have the value of @"BAR" myMethodThatTakesAnNSString(FOOLITERAL); I expect that I am just missing the obvious somehow, but I cannot seem to find the right preprocessor voodoo to get what I need. 回答1:

问题 Perhaps it is not good programming practice, but is it possible to define a for loop macro? For example, #define loop(n) for(int ii = 0; ii < n; ++ ii) works perfectly well, but does not give you the ability to change the variable name ii . It can be used: loop(5) { cout << "hi" << " " << "the value of ii is:" << " " << ii << endl; } But there is no choice of the name/symbol ii . Is it possible to do something like this? loop(symbol_name, n) where the programmer inserts a symbol name into "

问题 What's the meaning of #line in the C language? Where would it be used? 回答1: It tells the compiler where the following line actually came from. It's usually only the C preprocessor that adds these, for example, when including a file, it tells the compiler (which is basically only seeing one stream of data) that we're looking at a different file. This may sound strange, but the preprocessor simply inserts the header files where you specify your includes, and the compiler works on the whole

问题 I'm trying to pass the parameters to macro SETBIT with another predefined macro like this: #define SETBIT(ADDRESS,BIT,N) {(N) ? (ADDRESS &= ~(1<<BIT)) : (ADDRESS |= (1<<BIT))} #define DAC_SYNC PORTB,3,POS SETBIT(DAC_SYNC); However I receiver error: macro SETBIT requires 3 parameters only 1 given There is an article with the following recommendations: to prevent misnesting of arithmetic operations: #define foo (a,b) or #define bar(x) lose((x)) But even though I still have an error. BTW,

来源： https://stackoverflow.com/questions/16576415/are-include-directives-processed-prior-to-macro-expansion-regardless-of-their-l

来源： https://stackoverflow.com/questions/4948877/array-initialization-macro