如何在Cim中将C风格的printf转换为C ++风格的cout

Question

I got hand over some legacy code and first I want to see if it's possible to change something like

printf("test %d\n", var);

into

std::cout << "test " << var << std::endl;

There are a lot of them and doing them manually is very time consuming. Is there a way to use vim to make this happen?

The furthest I get is

:%s/printf(\(.*\), \(.*\));/std::cout << \1 << \2 << std::endl;/g

but this only gets me

std::cout << "test %d\n" << var << std::endl;

I can apply clang format to the code so in printf I can guarantee there is always a space after a comma. In this example the space is between comma and var.

Ideally this vim command would be able to detect the percentage sign to know how many variables is in there and also detect \\n to know when to replace it with std::endl. Please advice.

Answer 1

You could try writing a macro.

First, A macro to convert the "%[insert here]" to the cout format.

:reg
--- Registers
"f   0f%2xmcf"f,dwdw`ci" << ^[pa << "^[

So what this does on macro key F, is

1. 0f% Go to the start of the line, find the first percentage sign (assumes % is not used in any other way, sadly)
2. 2xmc Delete the %d part, then store the current position in the line on marker c.
3. f"f, Find the end of the string, then find the first comma.
4. dwdw Delete the comma, then delete the variable name. This will allow the variable to be stored so it can be pasted later.
5. [backtick]c Go to the stored position in the line at mark c.
6. i" << ^[ Insert " << into the string and escape back to command mode.
7. pa << "^[ Paste the stored variable name, then insert the string << " .

So, the end result is this

printf("test %d\n", var); // Before
printf("test " << var << "\n"); // After

On another macro key R, simply replay macro F multiple times (say 100). It won't finish the macro if you have less than 100 variables because f% will fail.

:reg
--- Registers
"r   100@f

So, an example would be

printf("test %d %d %d %d %d\n", var, var1, var2, var3, var4); // Before
printf("test " << var << " " << var1 << " " << var2 << " " << var3 << " " << var4 << "\n"); // After

Now write a macro C to convert the start and end to C++!

:reg
--- Registers
"c   0trcwcout << ^[f(x$F)x

1. 0tr Go to start of line, and find UNTIL first r. This is in case of different indentation levels. We will be positioned at the p in printf .
2. cwcout << ^[ Change printf to cout << and escape to command mode.
3. f(x Find the ( in printf( and delete it.
4. $F)x Go to the end of the line, find the last ) and delete it.

This gives:

printf("test " << var << " " << var1 << " " << var2 << " " << var3 << " " << var4 << "\n"); // Before
cout << "test " << var << " " << var1 << " " << var2 << " " << var3 << " " << var4 << "\n"; // After

To tie it all together, make another macro T, which finds printf , runs macro C, then runs macro F. It is done in this order so that if an early part fails, the rest of the command won't run.

:reg
--- Registers
"t   /printf^M@c@r

Running this macro T 3 times does the following:

// Before
printf("test %d\n", var);
printf("test %d %d %d %d %d\n", var, var1, var2, var3, var4);
printf("test %d %d\n", var, var2);

// After
cout << "test " << var << "\n";
cout << "test " << var << " " << var1 << " " << var2 << " " << var3 << " " << var4 << "\n";
cout << "test " << var << " " << var2 << "\n";

This solution isn't perfect, it assumes there that percentages aren't used in any other format, and it has to be manually repeated for each printf (spamming @@ ). I hope it is at least useful, and shows the power of vim.

Answer 2

This is actually harder than it seems, especially if you have more complicated formats with with field-width specifiers or other such things, or worse the "%[" format. It is also complicated if the arguments to the format string are more complicated than simple variables or literals, for example if you have function calls.

However for simple formatting strings and arguments, such as the printf call shown in the question, it's not that hard to do in a script language. You get the formatting string and put it in a string variable, then you get all the arguments and put them as a string in another variable. Split the argument string on comma, and you have a list of the arguments.

Then iterate over the formatting string, and when you hit a '%' character, which is not followed by another '%' character, then print the formatting string up to that point and get the first argument from the list of arguments. Then continue to scan the formatting string and get each corresponding argument from its list when you hit a formatting sequence.

Answer 3

我匹配了%d\\n"在捕获之外并在替换中添加了空格和结束”。

:%s/printf(\(.*\) \S\+, \(.*\));/std::cout << \1 " << \2 << std::endl;/g

Answer 4

Answer needs to be more generic! So, we can use a vim function to detect both % characters like %d%f%s as well as escape sequence characters like \\n\\r .

Just a sample psuedocode.

        :function ConvertToCPP()
        :let a = getline('.')
        : while a doesn't contain % or \ characters
        :    take any number of  \n and put as endl at end of line, if any
        :   remove off the % characters.
        : put << in between each of them
        :   call setline ('.', a)
        :return 1
        :end function

Noe, you can call this function with range, such that it will easily replace them all.

    :10,20ConvertToCPP()