About ARM's B,BL jump instruction

. text:00001260 A3 EB BL sub_24f4

Current address + (offset << 2 + 8) = Destination Address

0x1260 + 0X4A3 * 4 + 8 = 24f4

Offset = (Destination address-current address-8) >> 2

offset = (0x24f4-0x1260-8)/4 = 0X4A3

Assuming that the address at the jump command is a, the address at the destination of the jump is B.
The B,BL instruction holds the offset address, which is calculated by:
1.b-(a+8). A+8 is the actual value of the PC is a+8 when the arm assembly line makes the instruction execute at a.
2. The value obtained in the first step is a multiple of 4, because arm's instruction is 4 aligned, that is, the lowest two bits are 00. The value is then shifted to the right by two bits.
3. Get the final offset

When executing:
1. Remove the offset
2. Move left two-bit
3. Add a PC, when the value of the PC is exactly the address value of the target, that is, the target address instruction entered the reference, the first line two is emptied
http://www.decell.org/article.asp?id=2

About ARM's B,BL jump instruction