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Home/ Questions/Q 8406597
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Editorial Team
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Editorial Team
Asked: June 9, 20262026-06-09T23:09:09+00:00 2026-06-09T23:09:09+00:00

I want to parser cpu info in Linux. I wrote such code: // Returns

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I want to parser cpu info in Linux. I wrote such code:

// Returns full data of the file in a string
std::string filedata = readFile("/proc/cpuinfo");

std::cmath results;
// In file that string looks like: 'model name : Intel ...'
std::regex reg("model name: *");
std::regex_search(filedata.c_str(), results, reg);

std::cout << results[0] << " " << results[1] << std::endl;

But it returns empty string. What’s wrong?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-09T23:09:11+00:00Added an answer on June 9, 2026 at 11:09 pm

    You didn’t specify any capture in your expression.

    Given the structure of /proc/cpuinfo, I’d probably prefer a line
    oriented input, using std::getline, rather than trying to do
    everything at once. So you’ld end up with something like:

    std::string line;
    while ( std::getline( input, line ) ) {
        static std::regex const procInfo( "model name\\s*: (.*)" );
        std::cmatch results;
        if ( std::regex_match( line, results, procInfo ) ) {
            std::cout << "???" << " " << results[1] << std::endl;
        }
    }
    

    It’s not clear to me what you wanted as output. Probably, you also
    have to capture the processor line as well, and output that at the
    start of the processor info line.

    The important things to note are:

    1. You need to accept varying amounts of white space: use "\\s*" for 0 or more, "\\s+" for one or more whitespace characters.

    2. You need to use parentheses to delimit what you want to capture.

    (FWIW: I’m actually basing my statements on boost::regex, since I
    don’t have access to std::regex. I think that they’re pretty similar,
    however, and that my statements above apply to both.)

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