Hi Matthias,

Quoting Matthias Brugger [off-list ref]:

On 16/05/17 23:56, Gustavo A. R. Silva wrote:

quoted

Hello everybody,

While looking into Coverity ID 1408721 I ran into the following  
piece of code at /drivers/pwm/pwm-mediatek.c:77:

 77static int mtk_pwm_config(struct pwm_chip *chip, struct pwm_device *pwm,
 78                          int duty_ns, int period_ns)
 79{
 80        struct mtk_pwm_chip *pc = to_mtk_pwm_chip(chip);
 81        struct clk *clk = pc->clks[MTK_CLK_PWM1 + pwm->hwpwm];
 82        u32 resolution, clkdiv = 0;
 83
 84        resolution = NSEC_PER_SEC / clk_get_rate(clk);
 85
 86        while (period_ns / resolution > 8191) {
 87                resolution *= 2;
 88                clkdiv++;
 89        }
 90
 91        if (clkdiv > 7)
 92                return -EINVAL;
 93
 94        mtk_pwm_writel(pc, pwm->hwpwm, PWMCON, BIT(15) | BIT(3)  
| clkdiv);
 95        mtk_pwm_writel(pc, pwm->hwpwm, PWMDWIDTH, period_ns /  
resolution);
 96        mtk_pwm_writel(pc, pwm->hwpwm, PWMTHRES, duty_ns / resolution);
 97
 98        return 0;
 99}

The issue here is that in case _clk_ is null, function  
clk_get_rate() at line 84 will return zero and a division by zero  
will occur.

So my question here is if there is any chance for _clk_ to be null  
at line 81, hence ending up triggering a division by zero at line 84?

No it can't, in the probe function will error out if one of the  
seven clocks are not found.

for (i = 0; i < MTK_CLK_MAX; i++) {
	pc->clks[i] = devm_clk_get(&pdev->dev, mtk_pwm_clk_name[i]);
	if (IS_ERR(pc->clks[i]))
		return PTR_ERR(pc->clks[i]);
}

It registers a pwm chip with five PWMs. When the config function is  
called one of the five PWMs is identified (with a value from 0-4)  
which correspondents to the MTK_CLK_PWM[1-5], so no bug here neither.

I get it.

Thank you very much for the clarification.
--
Gustavo A. R. Silva