linux中,MEM,流量都可以比较好的得到,但CPU的使用率有点麻烦,经过一番测试,把top的源码稍改了一下,可以得到,源码如下:
#include
#include
#include
#include
#include
#define NCPUSTATES 5
static long cp_time[NCPUSTATES];
static long cp_old[NCPUSTATES];
static long cp_diff[NCPUSTATES];
int cpu_states[NCPUSTATES];
char buffer[4096+1];
static inline char *
skip_token(const char *p)
{
while (isspace(*p)) p++;
while (*p && !isspace(*p)) p++;
return (char *)p;
}
long percentages(cnt, out, new, old, diffs)
int cnt;
int *out;
register long *new;
register long *old;
long *diffs;
{
register int i;
register long change;
register long total_change;
register long *dp;
long half_total;
/* initialization */
total_change = 0;
dp = diffs;
/* calculate changes for each state and the overall change */
for (i = 0; i < cnt; i++)
{
if ((change = *new - *old) < 0)
{
/* this only happens when the counter wraps */
change = (int)
((unsigned long)*new-(unsigned long)*old);
}
total_change += (*dp++ = change);
*old++ = *new++;
}
/* avoid divide by zero potential */
if (total_change == 0)
{
total_change = 1;
}
/* calculate percentages based on overall change, rounding up */
half_total = total_change / 2l;
half_total = 0;
for (i = 0; i < cnt; i++)
{
//printf("dd %ld %ld\n",(*diffs* 1000 + half_total),total_change);
*out++ = (int)((*diffs++ * 1000 + half_total) / total_change);
}
/* return the total in case the caller wants to use it */
return(total_change);
}
int main(){
int fd,len;
char *p;
int i;
for(i = 0; i < 4; i++){
cpu_states[i] = 0;
cp_diff[i] = 0;
}
while(1){
fd = open("/proc/stat", O_RDONLY);
len = read(fd, buffer, sizeof(buffer)-1);
close(fd);
buffer[len] = '\0';
p = skip_token(buffer); /* "cpu" */
cp_time[0] = strtoul(p, &p, 0);
cp_time[1] = strtoul(p, &p, 0);
cp_time[2] = strtoul(p, &p, 0);
cp_time[3] = strtoul(p, &p, 0);
cp_time[4] = strtoul(p,&p,0);
//printf("f %ld %ld %ld %ld\n",cp_time[0],cp_time[1],cp_time[2],cp_time[3]);
percentages(NCPUSTATES, cpu_states, cp_time, cp_old, cp_diff);
printf("cpu used:%4.1f nice:%4.1f sys:%4.1f idle:%4.1f
iowait:%4.1f\n",cpu_states[0]/10.0,cpu_states[1]/10.0,cpu_states[2]/10.0,cpu_states[3]/10.0,cpu_states[4]/10.0);
sleep(3);
}
}
阅读(868) | 评论(0) | 转发(0) |