After working for more than 10 years, today a code caught my eye, I am unable to understand the function name defined inside a function gets printed in the output/log without being passed as an argument in macro or being defined as a global variable. Please help me understanding the internal. Please see the screenshot for the reference.
/*...*/
#include <stdio.h>
#define x printf("%s", f);
int main() {
char *f = "MAIN";
printf("Hello World");
x;
return 0;
}
Output:
Hello WorldMAIN
>Solution :
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 note that if there is no f declared when this macro is used, you will see a compiler error. If f is declared, but is not a char *, you should see compiler warnings.
Some preprocessor macro best practices include (but are not limited to) using capitalized names, as x by convention looks like a variable or function name; and being careful about what syntactically significant symbols (in this case ;) you include in your macro text.
Hopefully this example was done for the sake of learning, because it is wholly unnecessary. Preprocessor macros wouldn’t exist if they didn’t serve a purpose, but beware they can easily obfuscate code.