From: Heilpern, M. <mar...@au...> - 2006-08-23 19:39:09
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I'm trying to tie a GPIO to an interrupt and though I can see the line getting set (using a logic analyzer and also /proc/gpio/GPIO29), the interrupt doesn't happen. Any ideas? =20 Here is the relevant code pieces for my module: =20 #define GPIO_INTERRUPT 29 ... pxa_gpio_mode(GPIO_INTERRUPT | GPIO_IN); ... if (request_irq(IRQ_GPIO(GPIO_INTERRUPT), autc_irq, SA_SHIRQ, "AES", (void *)FPS_TYPE)) { /* if we reach this point, the ISR could not be installed */ err("ISR cannot be installed :("); autc_ps_release_hw(); return -1; } ... set_irq_type(IRQ_GPIO(GPIO_INTERRUPT), IRQT_RISING); ... // do something that will cause GPIO29 to rise ... ... ... =20 =20 =20 =20 I can see the GPIO is set as an input, and currently set, using proc_gpio. My trivial ISR performs a printk() so I'll know if it's been hit, but it never shows up in my system log. "cat /proc/interrupts" also shows 0 as the run count for this interrupt: 52: 0 AES =20 =20 I haven't done anything to explicitly disable the interrupt vector (and I haven't done anything to explicitly enable it either). =20 _______________________________________________________ NOTE: The information in this message is intended for the personal and = confidential use of the designated recipient(s) named above. To the extent the = recipient(s) is/are bound by a non-disclosure agreement, or other agreement that contains an = obligation of confidentiality, with AuthenTec, then this message and/or any = attachments shall be considered confidential information and subject to the confidentiality = terms of that agreement. If the reader of this message is not the intended recipient = named above, you are notified that you have received this document in error, and any = review, dissemination, distribution or copying of this message is strictly prohibited. If you = have received this document in error, please delete the original message and notify the = sender immediately. Thank you. AuthenTec, Inc. http://www.authentec.com |