How do you determine the amount of Linux system RAM in C++?

问题

I just wrote the following C++ function to programatically determine how much RAM a system has installed. It works, but it seems to me that there should be a simpler way to do this. Can someone tell me if I'm missing something?

getRAM()
{
    FILE* stream = popen( "head -n1 /proc/meminfo", "r" );
    std::ostringstream output;
    int bufsize = 128;

    while( !feof( stream ) && !ferror( stream ))
    {
        char buf[bufsize];
        int bytesRead = fread( buf, 1, bufsize, stream );
        output.write( buf, bytesRead );
    }
    std::string result = output.str();

    std::string label, ram;
    std::istringstream iss(result);
    iss >> label;
    iss >> ram;

    return ram;
}

First, I'm using popen("head -n1 /proc/meminfo") to get the first line of the meminfo file from the system. The output of that command looks like

MemTotal: 775280 kB

Once I've got that output in an istringstream, it's simple to tokenize it to get at the information I want. My question is, is there a simpler way to read in the output of this command? Is there a standard C++ library call to read in the amount of system RAM?

回答1:

On Linux, you can use the function sysinfo which sets values in the following struct:

   #include <sys/sysinfo.h>

   int sysinfo(struct sysinfo *info);

   struct sysinfo {
       long uptime;             /* Seconds since boot */
       unsigned long loads[3];  /* 1, 5, and 15 minute load averages */
       unsigned long totalram;  /* Total usable main memory size */
       unsigned long freeram;   /* Available memory size */
       unsigned long sharedram; /* Amount of shared memory */
       unsigned long bufferram; /* Memory used by buffers */
       unsigned long totalswap; /* Total swap space size */
       unsigned long freeswap;  /* swap space still available */
       unsigned short procs;    /* Number of current processes */
       unsigned long totalhigh; /* Total high memory size */
       unsigned long freehigh;  /* Available high memory size */
       unsigned int mem_unit;   /* Memory unit size in bytes */
       char _f[20-2*sizeof(long)-sizeof(int)]; /* Padding for libc5 */
   };

If you want to do it solely using functions of C++ (i would stick to sysinfo), i recommend taking a C++ approach using std::ifstream and std::string:

unsigned long get_mem_total() {
    std::string token;
    std::ifstream file("/proc/meminfo");
    while(file >> token) {
        if(token == "MemTotal:") {
            unsigned long mem;
            if(file >> mem) {
                return mem;
            } else {
                return 0;       
            }
        }
        // ignore rest of the line
        file.ignore(std::numeric_limits<std::streamsize>::max(), '\n');
    }
    return 0; // nothing found
}

回答2:

There's no need to use popen(), you can just read the file yourself. Also, if there first line isn't what you're looking for, you'll fail, since head -n1 only reads the first line and then exits. I'm not sure why you're mixing C and C++ I/O like that; it's perfectly OK, but you should probably opt to go all C or all C++. I'd probably do it something like this:

int GetRamInKB(void)
{
    FILE *meminfo = fopen("/proc/meminfo", "r");
    if(meminfo == NULL)
        ... // handle error

    char line[256];
    while(fgets(line, sizeof(line), meminfo))
    {
        int ram;
        if(sscanf(line, "MemTotal: %d kB", &ram) == 1)
        {
            fclose(meminfo);
            return ram;
        }
    }

    // If we got here, then we couldn't find the proper line in the meminfo file:
    // do something appropriate like return an error code, throw an exception, etc.
    fclose(meminfo);
    return -1;
}

回答3:

Remember /proc/meminfo is just a file. Open the file, read the first line, close the file. Voilá!

回答4:

Even top (from procps) parses /proc/meminfo, see here.

来源：https://stackoverflow.com/questions/349889/how-do-you-determine-the-amount-of-linux-system-ram-in-c