c++ - 总 RAM 大小 linux sysinfo 与/proc/meminfo

Question

 struct sysinfo sys_info;
 int32_t total_ram = 0;    
 if (sysinfo(&sys_info) != -1)
   total_ram = (sys_info.totalram * sys_info.mem_unit)/1024;

上面代码中total_ram的值为3671864。但是/proc/meminfo显示了不同的值。

cat /proc/meminfo | grep MemTotal
MemTotal:       16255004 kB

为什么它们不同？在 Linux 中获取物理 RAM 大小的正确方法是什么？

Answer 1

这是由于溢出造成的。当涉及到超过 40 亿的数量(例如 4GB+ RAM)时，请确保使用 64bit+ 类型:

 struct sysinfo sys_info;
 int32_t total_ram = 0;    
 if (sysinfo(&sys_info) != -1)
   total_ram = ((uint64_t) sys_info.totalram * sys_info.mem_unit)/1024;

这是一个独立的示例:

#include <stdint.h>
#include <stdio.h>
#include <sys/sysinfo.h>

int main() {
  struct sysinfo sys_info;
  int32_t before, after;
  if (sysinfo(&sys_info) == -1) return 1;

  before = (sys_info.totalram * sys_info.mem_unit)/1024;
  after = ((uint64_t)sys_info.totalram * sys_info.mem_unit)/1024;
  printf("32bit intermediate calculations gives %d\n", before);
  printf("64bit intermediate calculations gives %d\n", after);
  return 0;
}

编译并运行时:

$ gcc foo.c -o foo -m32 -Wall -Werror -ansi -pedantic && ./foo
32bit intermediate calculations gives 2994988
64bit intermediate calculations gives 61715244