![]() It then attempts to match it to a previously recorded IR signalĬode is public domain, check out and adafruit. SIGSELECT is a bit harder, but the member is named ' signo ' which implies that this is a signal number. So U32 would be a 32-bit unsigned integer, what is known as uint32t in C99 but has not yet been standardized in C++. This sketch/program uses the Arduno and a PNA4602 to decode IR received. A guess would be that the Uxx types are 'unsigned integers', of the specified bit widths. Okay then, here’s the code ! /* Raw IR commander Each uint16t is two bytes, so at NUMPULSES50 you’re using 50 2 2 200 bytes, at 70 it’s 280 bytes. pulses is a 2 dimensional array, so it’s got NUMPULSES arrays of two uint16t’s. Hmm, so well within the range - i tried to compile the code twice, once each with the 2 values, and they both resulted in Binary sketch size: 4,908 bytes (of a 32,256 byte maximum) Yes: uint16t is a datatype that’s unsigned and is 16 bits wide. Posting all the code and what board you’re using would help. All problems in computer science can be solved by another level of indirection, except for the problem of too many layers of indirection. on the 16-bit compiler you may need lu for the last one as it will be probably typedef of unsigned long. you can print them all using u format PS. Depending on what else is in the program, that might be pushing past the available memory for your board. this is unsigned integer type with 8,16,32 bits in it. Each uint16_t is two bytes, so at NUMPULSES=50 you’re using 50 22 = 200 bytes, at 70 it’s 280 bytes. Pulses is a 2 dimensional array, so it’s got NUMPULSES arrays of two uint16_t’s. Uint16_t is a datatype that’s unsigned and is 16 bits wide.
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