delay timer的wrap around

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TF-A链接：

https://github.com/ARM-software/arm-trusted-firmware

 

在阅读TF-A源代码时，看到其udelay是实现如下：

https://github.com/ARM-software/arm-trusted-firmware/blob/master/drivers/delay_timer/delay_timer.c

 

/***********************************************************
 * Delay for the given number of microseconds. The driver must
 * be initialized before calling this function.
 ***********************************************************/
void udelay(uint32_t usec)
{
    assert((timer_ops != NULL) &&
        (timer_ops->clk_mult != 0U) &&
        (timer_ops->clk_div != 0U) &&
        (timer_ops->get_timer_value != NULL));

    uint32_t start, delta, total_delta;

    assert(usec < (UINT32_MAX / timer_ops->clk_div));

    start = timer_ops->get_timer_value();

    /* Add an extra tick to avoid delaying less than requested. */
    total_delta =
        div_round_up(usec * timer_ops->clk_div,
                        timer_ops->clk_mult) + 1U;

    do {
        /*
         * If the timer value wraps around, the subtraction will
         * overflow and it will still give the correct result.
         */
        delta = start - timer_ops->get_timer_value(); /* Decreasing counter */

    } while (delta < total_delta);
}

 

第16行的get_timer_value返回的是一个从0xFFFFFFFF到0的递减的数，减到0后，再往下减的话，会重新变成0xFFFFFFFF。

其中第25到26行注释说即使发生了wraps around，也可以保证delta的值是正确的。下面我们看看是什么原理。

为此，下面是一个简单的模拟程序，看看发生wraps around后，两个数字相减的结果，假设下面的a，b，c的范围都是从0到0xFF，a表示第一次读到的数，b表示第二次读到的数。

 

#include <stdio.h>

int main(int argc, const char *argv[])
{
        unsigned char a, b, c;

        a = 0x20;
        b = 0x00;
        c = a - b;
        printf("a(0x%x) - b(0x%x) = %x\n", a, b, c);

        a = 0x00;
        b = 0xE0;
        c = a - b;
        printf("a(0x%x) - b(0x%x) = %x\n", a, b, c);

        a = 0x00;
        b = 0x20;
        c = a - b;
        printf("a(0x%x) - b(0x%x) = %x\n", a, b, c);

        a = 0xF0;
        b = 0x10;
        c = a - b;
        printf("a(0x%x) - b(0x%x) = %x\n", a, b, c);

        return ;
}

 

下面是运行结果：

a(0x20) - b(0x0) =
a(0x0) - b(0xe0) =
a(0x0) - b(0x20) = e0
a(0xf0) - b(0x10) = e0

可以看到，如果是以无符号数输出的话，上面的大数减小数和小数减大数的结果是一样的。

原因是负数在计算机中是以补码的形式存放，关于补码的介绍可以参考：https://baike.baidu.com/item/%E8%A1%A5%E7%A0%81/6854613?fr=aladdin

 

可以直观的理解，在模为10的情况下，一个数加上A，跟减去（10-A）的结果一样：比如 （3 + 8）%10 = 1，而(3-（10-8））%10 = 1，二者一样。

 

或者可以手动计算一下：0 - 0xe0，理论上应该等于-0xe0，那么这个数字在计算机中如何表示呢？也就是-0xe0的补码是多少？根据负数求补码的方法，其绝对值各位取反然后再加1：

-0xe0 --> 0xe(绝对值) --> 0b11100000(二进制表示) --> 0b00011111(取反) --> 0b00100000(加一) --> 0x20，即在模为0x100的情况下，-0xe0跟0x20是一回事。

 

完。