Forum Discussion
Altera_Forum
Honored Contributor
14 years agoAdd the values together and compare against the size of the available memory block!
17k + 12k = 29k I suspect that is a 32k on-chip memory block 28k + 4060k = 4088k Looks like 4MB (4096k) The other few k is probably bss and other pad space. (The standard linker script probably puts a gap between the code and data for mmu page permissions)