arm cortex m0 lpc1114 register configuration

The various data registers and control registers inside a 32-bit microcontroller are all 32 bits. Similarly, the data and control registers inside an 8-bit microcontroller are all 8 bits.

For example:

The definition of "interrupt control register" IE of AT89C51 microcontroller is shown in the figure below:

The definition of LPC1114's "AHB bus clock control register" SYSAHBCLKCTRL is shown in the figure below:

The above two control registers, one is 8-bit and the other is 32-bit. The similarity between them is that each bit determines a task. For example, writing 1 to bit4 of the IE register of the AT89C51 microcontroller can turn on the serial port interrupt, and writing 0 can turn off the serial port interrupt. Writing 1 to bit6 of the AHBCLKCTRL register of the LPC1114 microcontroller means turning on the GPIO working clock, and writing 0 means turning off the GPIO working clock.

The IE register mentioned above can be used to enable the general interrupt with IE=0x80, or directly write EA=1 to enable the general interrupt. The way to enable with EA=1 is "bit operation". Compared with directly writing the register value, "bit operation" will change the value of the entire register, while "bit operation" will not change other values ​​in the register.

The registers of the LPC1114 microcontroller do not support "bit operations". In order to operate a certain bit without affecting the values ​​of other bits, we need to use the logical "or" and "and" operations of the C language.

For example:

Write 1 to bit6 of the SYSAHBCLKCTRL register:

LPC_SYSCON->SYSAHBCLKCTRL |= (1<<6);

Write 0 to bit6 of the SYSAHBCLKCTRL register:

LPC_SYSCON->SYSAHBCLKCTRL &= ~(1<<6);

In the header file lpc11xx.h, each register is defined by the structure of each module, so when we want to write a value to a register, we need to use the symbol "->" to assign a value to a member variable of the structure. In the above formula, the SYSAHBCLKCTTL register is located in the structure LPC_SYSCON, so when assigning a value to the register, we need to write it like this.

1<<6 means 1 is shifted to the left by 6, that is:

The 32-bit number 1 is represented in binary as 000000000000000000000000000000001

The 32-bit number 1 shifted left 6 times becomes 000000000000000000000000001000000

The left-shifted data is "OR"ed with the value in SYSAHBCLKCTRL. The logic of "OR" is 0, "OR" any number is any number, and 1, "OR" any number is 1, so the result is only bit6 set to 1.

Similarly, we can analyze the statement that writes 0 to bit6. These are all basic knowledge of C language. Please believe that experts do not have special skills, but have mastered a solid foundation.

Special note: The two statements for writing registers above may seem complicated at first glance, but they are actually very simple! When we want to write a value to a certain bit of a register in the future, for example, to write 1 to bit n of the BB register of the AA module, we can apply the above formula, that is:

AA->BB |=(1<

Similarly, to write 0 to bit n, that is:

AA->BB&=~(1<

Dear! Congratulations, you have learned most of it.