macro output doesn't work - c++

Question

I'm learning c++,The macro behavior not as expected.
1 #include<cstdlib>
2 #include<iostream>
3 #include<cstring>
4 #define die(x) std::cout << x << std::endl ; exit(-1)
5 const char *help = "Usage: coffee --help --version";
6 const char *version = "alpha";
7 int main(int argc,char **argv)
8 {
9 if(argc<2||!strcmp(argv[1],"--help"))
10 die(help);
11 if(!strcmp(argv[1],"--version"))
12 die(version);
13
14 return 0;
15
16 }
g++ -o sample ./*
./sample --help
Output:Usage: coffee --help --version
./sample --version
Output:
I'm confused why --version didn't output string alpha.

Answer 1

When std::cout << x << std::endl ; exit(-1) is expanded by the macro preoprocessor in these two lines
9 if(argc<2||!strcmp(argv[1],"--help"))
10 die(help);
the resulting code is:
if(argc<2||!strcmp(argv[1],"--help"))
std::cout << help << std::endl;
exit(-1);
Which is probably not what you wanted;
The common trick for "multistatement macros" is to use do { ... } while(0) around the statements you want to have in a macro.
You can use gcc -E or cl -E to get the output from the C preprocessor, so you can see what the compiler ACTUALLY sees.
Edit: I should point out that I personnaly would prefer, in this case, a "die(msg) function" rather than fixing up the macro. Then you can, for example, set a breakpoing in die() and find out how you got there when something isn't working right! You can't set a breakoint in a macro.

Answer 2

Just try to brutally replace the body of the macro and you will see why:
if(argc<2||!strcmp(argv[1],"--help"))
die(help);
becomes:
if(argc<2||!strcmp(argv[1],"--help"))
std::cout << help << std::endl;
exit(-1);
With no braces { } for the if statement, the body is made just of one instruction so the exit(-1) is always executed.
You would have discovered if by using an if / else if instead that and if / if couple since the second else if would have missed its parent.

Answer 3

You forgot { }. Expand the macro manually and you'll see the result:
if(argc<2||!strcmp(argv[1],"--help"))
std::cout << help << std::endl ; exit(-1);
i.e.
if(argc<2||!strcmp(argv[1],"--help"))
std::cout << help << std::endl ;
exit(-1);

Answer 4

Your code after substitution of macro
if(argc<2||!strcmp(argv[1],"--help"))
std::cout << help << std::endl ; exit(-1) ; //<-- this exit will work always.
if(!strcmp(argv[1],"--version"))
std::cout << version << std::endl ; exit(-1) ;
correct way:
#define die(x) do {std::cout << x << std::endl ; exit(-1); } while(false);