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pwm: lpss-platform: Convert to platform remove callback returning void
The .remove() callback for a platform driver returns an int which makes many driver authors wrongly assume it's possible to do error handling by returning an error code. However the value returned is (mostly) ignored and this typically results in resource leaks. To improve here there is a quest to make the remove callback return void. In the first step of this quest all drivers are converted to .remove_new() which already returns void. Trivially convert this driver from always returning zero in the remove callback to the void returning variant. Signed-off-by: Uwe Kleine-König <u.kleine-koenig@pengutronix.de> Signed-off-by: Thierry Reding <thierry.reding@gmail.com>
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@ -62,10 +62,9 @@ static int pwm_lpss_probe_platform(struct platform_device *pdev)
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return 0;
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}
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static int pwm_lpss_remove_platform(struct platform_device *pdev)
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static void pwm_lpss_remove_platform(struct platform_device *pdev)
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{
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pm_runtime_disable(&pdev->dev);
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return 0;
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}
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static const struct acpi_device_id pwm_lpss_acpi_match[] = {
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@ -83,7 +82,7 @@ static struct platform_driver pwm_lpss_driver_platform = {
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.acpi_match_table = pwm_lpss_acpi_match,
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},
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.probe = pwm_lpss_probe_platform,
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.remove = pwm_lpss_remove_platform,
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.remove_new = pwm_lpss_remove_platform,
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};
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module_platform_driver(pwm_lpss_driver_platform);
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