c-preprocessor

问题 I have been trying to implement a function macro in C that prepends "DEBUG: ", to the argument, and passes its arguments to printf: #define DBG(format, ...) printf("DEBUG: " #format "\n", __VA_ARGS__) This gives me this error in gcc: src/include/debug.h:4:70: error: expected expression before ‘)’ token #define DBG(format, ...) printf("DEBUG: " #format "\n", __VA_ARGS__) ^ Supposedly, it should stringise format, and pass its variable arguments to printf, but so far I can't get past this error.

问题 I'd like to see all macros that are defined by the invocation of the compiler I'm using. Is there any way to do this? I have seen in the manual it says you can use cpp -dM but this doesn't work for me. Perhaps I'm doing something wrong? When I run: cpp -dM I get no output at all from the preprocessor. If I try adding -dM as an option on gcc, I don't notice any difference. 回答1: You can use: gcc -dM -E - < /dev/null Note that you can also get the compiler macros in addition with this command:

问题 This question already has answers here : Closed 8 years ago . Possible Duplicate: C preprocessor: using #if inside #define? Is there any trick to have preprocessor directives inside rhs of define ? The problem is, preprocessor folds all rhs into one long line. But maybe there is a trick ? Example of what I'd want in the rhs is #define MY_CHECK \ #ifndef MY_DEF \ # error MY_DEF not defined \ #endif ? The purpose is a shortness: to have 1-line shortcut instead of multiline sequence of checks.

问题 This question already has answers here : A riddle (in C) (4 answers) Closed 2 years ago . In the code shown below, nothing gets printed, which means the condition in the for loop fails. What could be the reason? I'm wondering because when I print TOTAL_ELEMENTS separately, it gives 5 , so naturally this must be 5-2=3 => -1<=3 , so it should print something. #define TOTAL_ELEMENTS (sizeof(array) / sizeof(array[0])) int array[] = { 23, 34, 12, 17, 204, 99, 16 }; int main() { int d; for (d = -1;

问题 Consider the following macro definitions and invocation: #define x x[0] #define y(arg) arg y(x) This invocation expands to x[0] (tested on Visual C++ 2010, g++ 4.1, mcpp 2.7.2, and Wave). Why? Specifically, why does it not expand to x[0][0] ? During macro replacement, A parameter in the replacement list...is replaced by the corresponding argument after all macros contained therein have been expanded. Before being substituted, each argument’s preprocessing tokens are completely macro replaced

问题 Here's a concrete example of what I want to do. Consider the string.Join function. Pre-.NET 4.0, there were only two overloads, both of which required a string[] parameter. As of .NET 4.0, there are new overloads taking more flexible parameter types, including IEnumerable<string> . I have a library which includes a Join function that does essentially what the .NET 4.0 string.Join function does. I was just wondering if I could make this function's implementation dependent on the .NET framework

问题 Why won't this work? 0. #define CONCAT(x, y) x ## y 1. 2. #define VAR_LINE(x) \ 3. int CONCAT(_anonymous, __LINE__) = x 4. 5. #define VAR_LINE2(x) \ 6. int _anonymous ## x = 1 7. 8. int main() 9. { 10. VAR_LINE(1); 11. VAR_LINE(1); 12. VAR_LINE(1); 13. VAR_LINE2(__LINE__); 14. } The result from the above macro expansion int _anonymous__LINE__ = 1; int _anonymous__LINE__ = 1; int _anonymous__LINE__ = 1; int _anonymous13 = 1; It would be convenient if I didn't have to write that __LINE__ macro

问题 What is a##b & #a ? #define f(a,b) a##b #define g(a) #a #define h(a) g(a) main() { printf("%s\n",h(f(1,2))); //how should I interpret this?? [line 1] printf("%s\n",g(f(1,2))); //and this? [line 2] } How does this program work? The output is 12 f(1, 2) now I understand how a##b & #a work. But why is the result different in the two cases (line 1 and line 2)? 回答1: The ## concatenates two tokens together. It can only be used in the preprocessor. f(1,2) becomes 1 ## 2 becomes 12 . The # operator

问题 This question already has answers here : The need for parentheses in macros in C [duplicate] (8 answers) Confusion with Macro expansion [duplicate] (3 answers) Closed 6 years ago . This question was asked to me in a mock interview...Really got surprised to find awkward answers... consider a macro: #define SQR(x) (x*x) Example 1: SQR(2) //prints 4 Example 2: If SQR(1+1) is given it doesn't sum (1+1) to 2 but rather ... SQR(1+1) //prints 3 Awkward right? What is the reason? How does this code

问题 Suppose we are given this input C code: #define Y 20 #define A(x) (10+x+Y) A(A(40)) gcc -E outputs like that (10+(10+40 +20)+20) . gcc -E -traditional-cpp outputs like that (10+(10+40+20)+20) . Why the default cpp inserts the space after 40 ? Where can I find the most detailed specification of the cpp that covers that logic ? 回答1: The C standard doesn't specify this behaviour, since the output of the preprocessing phase is simply a stream of tokens and whitespace. Serializing the stream of