How to write preprocessor definition that skips lines of code in C++?

Question

Let's say this is a preprocessor definition before function f():

#define write std::cout << "test";
write
void f()
{
    //...
}

and this is result of that macro:

std::cout << "test"
void f()
{
    //...
}

How do I write that macro so it will skip function and also insert some code behind the function so that the result will be something like this:

std::cout << "test";
void f()
{
    //...
}
std::cout << "test";

You know what I mean: a macro (or something else) that skips some code and inserts multiple lines.

James McNellis · Accepted Answer · 2011-12-31 06:42:58Z

6

You can pass the function itself as an argument to the macro:

#define write(...)       \
    std::cout << "test"; \
    __VA_ARGS__          \
    std::cout << "test";

write(
void f()
{
    //...
})

This particular example, of course, is ill-formed because there are statements outside of functions.

If you are actually interested in printing text at the beginning and end of a function, your best bet is to create a class that prints the text in its constructor and destructor, and declare an instance of that type at the beginning of the function.

answered Dec 31, 2011 at 6:42

James McNellis

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3 Comments

Vinayak Garg Over a year ago

+1 A nice solution but one that should be avoided. class solution recommended instead.

Sergey Kalinichenko Over a year ago

+1 A class-based constructor/destructor solution is the way to go: you could build elaborate things with it. We used it for tracing method entries and exits, and even to capture crash stacks in production environments.

codekiddy Over a year ago

Oh thanks, that's awsome :D, what do you mean by "create class option" will I put that object(of that class) at the beginig inside a function or outside? thanks. how does that work :)

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