C preprocessor macros simply do text replacement. They have no semantic awareness of your program. This: #include <stdio.h> #define x printf(“%s”, f); int main() { char* f = “MAIN”; printf (“Hello World”); x; return 0; } Becomes: #include <stdio.h> int main() { char* f = “MAIN”; printf (“Hello World”); printf(“%s”, f);; return 0; } Please … Read more
Introduction #define is a preprocessor directive in the C programming language that allows for the definition of macros. It is used to replace a function name with a predefined value. This is useful for creating constants, as well as for creating short-hand versions of commonly used functions. By using #define, the programmer can save time … Read more
Absolutely nothing. The pre-processor is simply replacing MY_IDENTIFIER with nothing wherever it encounters it. 1 solved What’s the meaning of #define without replacement in C/C++? [duplicate]
Make the arrays static so they will not be allocated on the stack. static const char fetch[] = “fetch”; static const char push[] = “push”; static const char quit[] = “quit”; solved How to use string literals without using arrays and #define preprocessor directive?
Here is the macro: #define IS_ALNUM(x) (((x)>=’a’ && (x) <= ‘z’)) ||((x)>=’A’ && (x) <= ‘Z’)) || (((x)>=’0′ && (x) <= ‘9’))) It tests if it is Between a and z Between A and Z Between 0 and 9 Quite simple 2 solved find the input is alphanumeric or not with using #define macro preprocessor … Read more
#define is a preprocessor directive, not a statement or declaration as defined by the C grammar (both of those are required to end with a semicolon). The rules for the syntax of each one are different. solved Why does #define not require a semicolon?
#define torsk break but I also wonder why would you want such weird thing… 1 solved Define ‘break’ as ‘torsk’
Preprocessor directives are interpreted before compiling your program. So VAR has no knowledge of a and b. See C preprocessor on Wikipedia. Instead, you could create a macro that takes parameters, like this: #define VAR(a,b) (a | b) …and use it like that: #if (VAR(a,b) != 0) You would have to adapt your program, since … Read more
Unfortunately, this is possible. But in no sense can I condone it. #include <iostream> namespace std { class not_actually_cout{}; template<typename T> not_actually_cout& operator<< (not_actually_cout& stream, const T & v) { std::cout << v << std::endl; return stream; } not_actually_cout not_actually_cout_instance; } #define cout not_actually_cout_instance int main(void) { cout << “why god why”; cout << “please … Read more
#define STR(x) printf(#x “=%d”,x) I misread the question..when you use it in printf..use #define STR(x) “%s = %d “,#x,x or #define STR(x) #x “=%d”,x 10 solved How to print variables with #define like #define STR(M) = value of M [closed]