Hi, could someone explain me how the if statement works in the snippet below? Despite the fact that I know how bitwise works I cannot understand when thewill be TRUE and when will be FALSE.Code:channel & (1 << i)
Code:void selectMuxPin(int channel) { for (int i = 0; i < MUX_CTRL_CHA; i++) { if (channel & (1 << i)) { nrf_gpio_pin_set(ctrlPins[i]); }else { nrf_gpio_pin_clear(ctrlPins[i]); } } }
It will result in true when the bit corresponding (2^i) in channel is set to 1 (where, '^' denotes "raised to the power of" and not XOR). Otherwise, false.
"Programming today is a race between software engineers striving to build bigger and better idiot-proof programs, and the Universe trying to produce bigger and better idiots. So far, the Universe is winning." - Rick Cook, The Wizardry Compiled
Like Zeus_ said... Here's a detailed explanation.Originally Posted by Nikosant03
When testing for true or false, any value different from 0 is TRUE. 0 is always FALSE. This expression will isolate bit i of the value from channel, all the other bits will be zero. If i is 3, for example, (1 << i) will be, in binary, 0b00000000000000000000000000001000. After the AND operation all bits are zero, except the 4th bit from right to left (bit 3), which will be a copy from the same bit from channel. If the 4th bit of channel is 0, the result will be 0, otherwise, 0x0004.
Got it?
Thank you guys, now I can see how it works. Basically anything besides 000 is a TRUE isn't it? I analyzed the function and the results are clear
MUX_CTRL_CHA = 3
1 = 001
channel = 0(000) then:
For i = 0 -> 001 << 0 = 001 , channel & (1 << i) -> 000 & 001 = 000 -> nrf_gpio_pin_clear 0
For i = 1 -> 001 << 1 = 010 , channel & (1 << i) -> 000 & 010 = 000 -> nrf_gpio_pin_set 0
For i = 2 -> 001 << 2 = 100 , channel & (1 << i) -> 000 & 100 = 000 -> nrf_gpio_pin_clear 0
channel = 1(001) then:
For i = 0 -> 001 << 0 = 001 , channel & (1 << i) -> 001 & 001 = 001 -> nrf_gpio_pin_set 1
For i = 1 -> 001 << 1 = 010 , channel & (1 << i) -> 001 & 010 = 000 -> nrf_gpio_pin_clear 0
For i = 2 -> 001 << 2 = 100 , channel & (1 << i) -> 001 & 100 = 000 -> nrf_gpio_pin_clear 0
channel = 2 (010) then:
For i = 0 -> 001 << 0 = 001 , channel & (1 << i) -> 010 & 001 = 000 -> nrf_gpio_pin_clear 0
For i = 1 -> 001 << 1 = 010 , channel & (1 << i) -> 010 & 010 = 010 -> nrf_gpio_pin_set 1
For i = 2 -> 001 << 2 = 100 , channel & (1 << i) -> 010 & 100 = 000 -> nrf_gpio_pin_clear 0
channel = 3 (011) then:
For i = 0 -> 001 << 0 = 001 , channel & (1 << i) -> 011 & 001 = 001 -> nrf_gpio_pin_set 1
For i = 1 -> 001 << 1 = 010 , channel & (1 << i) -> 011 & 010 = 010 -> nrf_gpio_pin_set 1
For i = 2 -> 001 << 2 = 100 , channel & (1 << i) -> 011 & 100 = 000 -> nrf_gpio_pin_clear 0
channel = 4 (100) then:
For i = 0 -> 001 << 0 = 001 , channel & (1 << i) -> 100 & 001 = 000 -> nrf_gpio_pin_clear 0
For i = 1 -> 001 << 1 = 010 , channel & (1 << i) -> 100 & 010 = 000 -> nrf_gpio_pin_clear 0
For i = 2 -> 001 << 2 = 100 , channel & (1 << i) -> 100 & 100 = 100 -> nrf_gpio_pin_set 1
channel = 5 (101) then:
For i = 0 -> 001 << 0 = 001 , channel & (1 << i) -> 101 & 001 = 001 -> nrf_gpio_pin_set 1
For i = 1 -> 001 << 1 = 010 , channel & (1 << i) -> 101 & 010 = 000 -> nrf_gpio_pin_clear 0
For i = 2 -> 001 << 2 = 100 , channel & (1 << i) -> 101 & 100 = 100 -> nrf_gpio_pin_set 1
channel = 6 (110) then:
For i = 0 -> 001 << 0 = 001 , channel & (1 << i) -> 110 & 001 = 000 -> nrf_gpio_pin_clear 0
For i = 1 -> 001 << 1 = 010 , channel & (1 << i) -> 110 & 010 = 010 -> nrf_gpio_pin_set 1
For i = 2 -> 001 << 2 = 100 , channel & (1 << i) -> 110 & 100 = 100 -> nrf_gpio_pin_set 1
channel = 7 (111) then:
For i = 0 -> 001 << 0 = 001 , channel & (1 << i) -> 111 & 001 = 001 -> nrf_gpio_pin_set 1
For i = 1 -> 001 << 1 = 010 , channel & (1 << i) -> 111 & 010 = 010 -> nrf_gpio_pin_set 1
For i = 2 -> 001 << 2 = 100 , channel & (1 << i) -> 111 & 100 = 100 -> nrf_gpio_pin_set 1
Yes.Basically anything besides 000 is a TRUE isn't it?
This is a basic bit test.
An easier example would be if I had a microcontroller with an 8-bit port, PORTA.Code:// A few simple bit tasks for u-controllers... // Test bit (1<<bit) & x // Set bit x |= (1<<bit); // Set multiple bits x |= (1<<bit1) | (1<<bit2) | (1<<bit3); // Clear bit x &= ~(1<<bit); // Clear multiple bits x &= ~((1<<bit1) | (1<<bit2) | (1<<bit3));
On the input for b3 (usually corresponds to the forth pin, because b0 is the first pin) I have a switch that pulls the pin to logic level high and we want a motor to start...
In the AVR world you see a lot of the "<<" notation to set and clear bits. I tend to do it with all different microcontroller family projects now days...Code:if ((1<<3) & PORTA) StartMotor();
This is so that you can see each individual bit.
You can see that the 0<< is kept in, just so a comment can be made as to why is was 0.Code:x = (1<<0) | //comment (1<<1) | //comment (0<<2) ; //comment
This is from a PIC, setting up the oscillator - Note that 0b is a non-standard but widely used way of writing binary literal and the _xx_POSITION is a defined number (0-7) in a header file for the XC8 compiler.
Code:OSCCON = (0b1110 << _OSCCON_IRCF_POSITION) | // 8 MHz or 32 MHz HF(see Section 5.2.2.1 ?HFINTOSC?) (0b00 << _OSCCON_SCS_POSITION) | // Clock determined by FOSC<2:0> in Configuration Word 1 (0 << _OSCCON_SPLLEN_POSITION); // 4x PLL Is enabled
Fact - Beethoven wrote his first symphony in C
Great answer!! Thank you for the sample code!!
Well... Bit inversion is missing:
Code:x ^= (1 << bit);
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Fact - Beethoven wrote his first symphony in C
