In C, string literals are concatenated automatically. For example, const char * s1 = “foo” “bar”; const char * s2 = “foobar”; s1 and s2 are the same string. So, for your problem, the answer (without token pasting) is #ifdef __TESTING #define IV_DOMAIN “example.org” #elif __LIVE_TESTING #define IV_DOMAIN “test.example.com” #else #define IV_DOMAIN “example.com” #endif #define … Read more
You can use various recursive macro techniques to do things with variadic macros. For example, you can define a NUM_ARGS macro that counts the number of arguments to a variadic macro: #define _NUM_ARGS(X100, X99, X98, X97, X96, X95, X94, X93, X92, X91, X90, X89, X88, X87, X86, X85, X84, X83, X82, X81, X80, X79, X78, … Read more
In C, string literals are concatenated automatically. For example, const char * s1 = “foo” “bar”; const char * s2 = “foobar”; s1 and s2 are the same string. So, for your problem, the answer (without token pasting) is #ifdef __TESTING #define IV_DOMAIN “example.org” #elif __LIVE_TESTING #define IV_DOMAIN “test.example.com” #else #define IV_DOMAIN “example.com” #endif #define … Read more
No, you can’t unstringify something.
Classical C preprocessor question…. #define STR_HELPER(x) #x #define STR(x) STR_HELPER(x) #define MAJOR_VER 2 #define MINOR_VER 6 #define MY_FILE “/home/user/.myapp” STR(MAJOR_VER) STR(MINOR_VER) The extra level of indirection will allow the preprocessor to expand the macros before they are converted to strings.
An occurrence of a parameter in a function-like macro, unless it is the operand of # or ##, is expanded before substituting it and rescanning the whole for further expansion. Because g‘s parameter is the operand of #, the argument is not expanded but instead immediately stringified (“f(1,2)”). Because h‘s parameter is not the operand … Read more
nothing too fancy: ## tells the preprocessor to concatenate the left and right sides see http://en.wikipedia.org/wiki/C_preprocessor#Token_concatenation
Like this: #include <stdio.h> #define QUOTE(str) #str #define EXPAND_AND_QUOTE(str) QUOTE(str) #define TEST thisisatest #define TESTE EXPAND_AND_QUOTE(TEST) int main() { printf(TESTE); } The reason is that when macro arguments are substituted into the macro body, they are expanded unless they appear with the # or ## preprocessor operators in that macro. So, str (with value TEST … Read more
see http://www.decompile.com/cpp/faq/file_and_line_error_string.htm specifically: #define STRINGIFY(x) #x #define TOSTRING(x) STRINGIFY(x) #define AT __FILE__ “:” TOSTRING(__LINE__) so your problem can be solved by doing sscanf(buf, “%” TOSTRING(MAX_LEN) “s”, val);
If you’re using c99 or c++0x there is the pragma operator, used as _Pragma(“argument”) which is equivalent to #pragma argument except it can be used in macros (see section 6.10.9 of the c99 standard, or 16.9 of the c++0x final committee draft) For example, #define STRINGIFY(a) #a #define DEFINE_DELETE_OBJECT(type) \ void delete_ ## type ## … Read more