How Many Registers Does Arm Have
The amount of registers depends on the ARM version. Co-ordinate to the ARM Reference Transmission, there are 30 general-purpose 32-bit registers, with the exception of ARMv6-Thousand and ARMv7-M based processors. The first 16 registers are accessible in user-level way, the additional registers are bachelor in privileged software execution (with the exception of ARMv6-Grand and ARMv7-K). In this tutorial series we volition work with the registers that are attainable in any privilege mode: r0-fifteen. These 16 registers can be split up into ii groups: general purpose and special purpose registers.
The following tabular array is only a quick glimpse into how the ARM registers could relate to those in Intel processors.
R0-R12: tin can be used during mutual operations to store temporary values, pointers (locations to retentiveness), etc. R0, for case, can exist referred as accumulator during the arithmetic operations or for storing the result of a previously chosen part. R7 becomes useful while working with syscalls as it stores the syscall number and R11 helps united states of america to keep track of boundaries on the stack serving equally the frame arrow (will be covered later). Moreover, the function calling convention on ARM specifies that the first four arguments of a role are stored in the registers r0-r3.
R13: SP (Stack Pointer). The Stack Pointer points to the elevation of the stack. The stack is an area of retentiveness used for office-specific storage, which is reclaimed when the function returns. The stack pointer is therefore used for allocating space on the stack, past subtracting the value (in bytes) we want to classify from the stack pointer. In other words, if we want to allocate a 32 bit value, we subtract four from the stack arrow.
R14: LR (Link Annals). When a function call is made, the Link Register gets updated with a memory address referencing the next instruction where the function was initiated from. Doing this allows the program render to the "parent" function that initiated the "child" function call after the "child" function is finished.
R15: PC (Plan Counter). The Program Counter is automatically incremented by the size of the instruction executed. This size is always 4 bytes in ARM state and 2 bytes in Pollex mode. When a branch instruction is being executed, the PC holds the destination address. During execution, PC stores the address of the current didactics plus eight (two ARM instructions) in ARM state, and the current instruction plus 4 (ii Thumb instructions) in Thumb(v1) country. This is unlike from x86 where PC ever points to the next instruction to be executed.
Let'southward look at how PC behaves in a debugger. Nosotros utilise the following program to store the address of pc into r0 and include two random instructions. Allow's run across what happens.
.section .text .global _start _start: mov r0, pc mov r1, #2 add together r2, r1, r1 bkpt
In GDB nosotros fix a breakpoint at _start and run it:
gef> br _start Breakpoint 1 at 0x8054 gef> run
Here is a screenshot of the output we see beginning:
$r0 0x00000000 $r1 0x00000000 $r2 0x00000000 $r3 0x00000000 $r4 0x00000000 $r5 0x00000000 $r6 0x00000000 $r7 0x00000000 $r8 0x00000000 $r9 0x00000000 $r10 0x00000000 $r11 0x00000000 $r12 0x00000000 $sp 0xbefff7e0 $lr 0x00000000 $pc 0x00008054 $cpsr 0x00000010 0x8054 <_start> mov r0, pc <- $pc 0x8058 <_start+4> mov r0, #2 0x805c <_start+8> add together r1, r0, r0 0x8060 <_start+12> bkpt 0x0000 0x8064 andeq r1, r0, r1, asr #x 0x8068 cmnvs r5, r0, lsl #2 0x806c tsteq r0, r2, ror #xviii 0x8070 andeq r0, r0, r11 0x8074 tsteq r8, r6, lsl #half dozen
Nosotros can see that PC holds the accost (0x8054) of the side by side instruction (mov r0, pc) that will be executed. Now let'southward execute the next instruction subsequently which R0 should hold the accost of PC (0x8054), correct?
$r0 0x0000805c $r1 0x00000000 $r2 0x00000000 $r3 0x00000000 $r4 0x00000000 $r5 0x00000000 $r6 0x00000000 $r7 0x00000000 $r8 0x00000000 $r9 0x00000000 $r10 0x00000000 $r11 0x00000000 $r12 0x00000000 $sp 0xbefff7e0 $lr 0x00000000 $pc 0x00008058 $cpsr 0x00000010 0x8058 <_start+4> mov r0, #ii <- $pc 0x805c <_start+viii> add r1, r0, r0 0x8060 <_start+12> bkpt 0x0000 0x8064 andeq r1, r0, r1, asr #x 0x8068 cmnvs r5, r0, lsl #ii 0x806c tsteq r0, r2, ror #18 0x8070 andeq r0, r0, r11 0x8074 tsteq r8, r6, lsl #half-dozen 0x8078 adfcssp f0, f0, #iv.0
…right? Wrong. Look at the accost in R0. While we expected R0 to comprise the previously read PC value (0x8054) it instead holds the value which is two instructions alee of the PC we previously read (0x805c). From this example you can see that when we directly read PC it follows the definition that PC points to the next instruction; merely when debugging, PC points two instructions ahead of the current PC value (0x8054 + eight = 0x805C). This is because older ARM processors always fetched two instructions ahead of the currently executed instructions. The reason ARM retains this definition is to ensure compatibility with earlier processors.
Source: https://azeria-labs.com/arm-data-types-and-registers-part-2/
Posted by: phelpstram1952.blogspot.com
0 Response to "How Many Registers Does Arm Have"
Post a Comment