memory size of C code

hi

after compilation of microblaze C code the log file says

---------------------------------------------- mb-size MB0/executable.elf text data bss dec hex filename

2780 657 4 3441 d71 MB0/executable.elf

----------------------------------------------

My first question is

- Is it mean that "memory size = 2780 + 3441 = 6221 = bytes? "

Function of C code is that add 300 integers using array declaration. The execution stops after adding 250 integers. This is undesirable. I think this problem is comming from limited BRAM size.

Second question is

- How can we maximally utilize memory (BRAM) so that we can perform more powerful operations ?

Reply to
Jack
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No, 2780 + 657 + 4 = 3441 bytes total in decimal, or D71 bytes total in hex.

Regards,

John McCaskill

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