I am debugging a C code which contains something related to Python. In a function:
void normalize(string &text) {
...
#ifdef _Python_CALL
newContentStr = contentStr;
#endif
#ifndef _Python_CALL
...
...
#endif
return 0;
}
I am using GDB to keep track of the code logic, and I found that after it reaches the line:
newContentStr = contentStr;
It then directly jumps to the last line in the function:
return 0;
Why is the code between the following is skipped?
#ifndef _Python_CALL
...
...
#endif
Also note that the first is "#ifdef" and the 2nd is "#ifndef". Does that make the skip?
CodePudding user response:
Judging from the code fragment you've shown, _Python_CALL is a macro name, possibly defined somewhere via #define _Python_CALL (or maybe by some other means, such as using a command line argument to the C compiler during compilation).
Then, line #ifdef _Python_CALL means that everything that follows it until the line #endif will be compiled (and thus executed in the compiled program) if and only if the macro name _Python_CALL is defined (the #ifdef means "if defined"). Since you claim that the line newContentStr = contentStr; was executed, we can assume that the macro name _Python_CALL was indeed defined during compilation.
Now, the line #ifndef _Python_CALL means that everything that follows it until the line #endif will be compiled (and executed) if and only if the macro name _Python_CALL is NOT defined. (Note the n in #ifndef, it means "if not defined"). But, as we already know (from the conclusion we made in the previous paragraph), this is not the case, because _Python_CALL is indeed defined. Thus, this block will not be compiled/executed.
On Cppreference, you can read more about C preprocessor, especially about #define and #ifdef / #ifndef directives, to gain deeper understanding.
CodePudding user response:
#ifndef is the opposite of #ifdef.
In your case, #ifdef is true, so it jump to return 0 directly and omit #ifndef block.
see this official doc
