Question: Consider a simple 8-bit configuration register, where each bit sets the configuration of 8 different pins ([7...0]), where setting a bit to 0 means configuring

Consider a simple 8-bit configuration register, where each bit sets the configuration of 8 different pins ([7...0]), where setting a bit to 0 means configuring input and setting 1 means output: char *ioDirConfigRegister = 0x20000000; // Location of configuration register in memory mapped IO If we want to set the 4th (0 indexed) pin to output, why is the following wrong? *ioDirConfigRegister = 1 << 4; Choice 1 of 5:We cannot use the shift operations here. We should use *ioDirConfigRegister = 0

Step by Step Solution

There are 3 Steps involved in it

1 Expert Approved Answer
Step: 1 Unlock

The code ioDirConfigRegister 1 4 is used to set the 4th pin 0indexed to output but it is incorrect for the following reason The code ioDirConfigRegist... View full answer

blur-text-image
Question Has Been Solved by an Expert!

Get step-by-step solutions from verified subject matter experts

Step: 2 Unlock
Step: 3 Unlock

Students Have Also Explored These Related Algorithms Questions!