如何以百分比百分比获取 C 程序中的 CPU 使用率

Question

I try to get a cpu usage with c program but the result does not come out in percentage %, what is wrong with this code? Maybe someone can clarify or offer me another way to recover the CPU used.

My code:

 int GetCPULoad() {
    int FileHandler;
    char FileBuffer[1024];
    float load;

    FileHandler = open("/proc/loadavg", O_RDONLY);
    if(FileHandler < 0) {
        return -1; }
    read(FileHandler, FileBuffer, sizeof(FileBuffer) - 1);
    sscanf(FileBuffer, "%f", &load);
    close(FileHandler);
    return (int)(load * 100);
}

maybe can i do that with /proc/stat file, any one know how to do that?

i have try another code

    fp = fopen ("/proc/stat", "r");
    if (fp) {
    long long unsigned int user,nice,system,idle;
    int i = fscanf(fp,"%*s %llu %llu %llu %llu",&user,&nice,&system,&idle);
    int total_cpu_usage1;
    total_cpu_usage1 = user + nice + system + idle;
    sprintf( http_buf,"<br>CPU usage: %d ", total_cpu_usage1); tcp_write(&tcpbuf, sock, http_buf, strlen(http_buf) );
    fclose(fp);
    }

but same result: CPU usage: 159838295

how can i get the persentage %?

Answer 1

The simplest way would be with multiple loops that have different ranges and different increments.

for (i = 1; i < 10; i++) {
    ...
}
for (i = 10; i <= 30; i += 10) {
    ...
}
for (i = 60; i <= 120; i += 30) {
    ...
}

If you really have to do it in a single loop:

for (i = 1; i <= 120; 
        i = i < 10 ? i + 1 : 
                     (i < 30 ? i + 10 : i + 30))

The last part of the for loop header increments i by different values depending on the range it's in.