How to caculate the addresses in LDF file

Question

Hi, 
 Here is a part of an LDF file which is copied from a refrence ADSP-21487 project: 
 
 seg_sigma_extdata { TYPE(PM RAM) START (0x000E4DD8) END ( 0x000E5260 ) WIDTH (48)} ss_block7_code { TYPE(SW RAM) START ( 0x001CF724 ) END (0x001CF72F) WIDTH (16)} ss_block8_data { TYPE(DM RAM) START (0x000E7B98) END (0x000E7BA3) WIDTH (32)} 
 
 If we want to reduce the length of 'seg_sigma_extdata' and enlarge 'ss_block7_code', how can we caculate new end address of 'seg_sigma_extdata' and new start address of 'ss_block7_code'? 
 Thanks.

Castor.Xu · Accepted Answer

Solved, found the answer in programming reference guide: 
 
 m = B + (3/2 (n – B)) + 1 where: • n is the first unused address after the end of 48-bit words • B is the base normal word 48-bit address of the internal memory block • m is the first 32-bit normal word address to use after the end of 48-bit words. 
 
 ...... 
 
 Example: Calculating a Starting Address for 32-Bit Addresses 
 Given a block of words in the range 0x90000 to 0x92694 (block 0), the next valid address is 0x92695. The number of 48-bit words (n) is: n = 0x92695 - 0x80000 = 0x12695. When 0x12695 is converted to decimal representation, the result is 75413. The base (B) normal word address of the internal memory block is 0x80000. The first 32-bit normal word address to use after the end of the 48-bit words is given by: m = 0x80000 + (3/2 (75413)) + 1 m = 0x80000 + 0x1B9E0 m = 0x80000 + 0x1B9E0 = 0x9B9E0 The first valid starting 32-bit address is 0x9B9E0.)