CODEWITHSUNDEEP
cgdbstack
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Reputation: 1847

gdb debugging what is the 8 byte gap for?

I'm debugging a function to learn more about the structure of a stack in memory. I am using gdb on the Ubuntu OS x86. My function consists of the following: 

function func(){
   long local1=0;
   printf("address of swap is %p\n",&local1);
}

In gdb I set a breakpoint inside the function and print out the frame values using info frame. I am able to get the address of the saved registers of ebp and eip which I presume holds the saved base pointer and return address respectively. I also print out the address of local1. So using these addresses I constructed the following stack: 

bffff03c --> eip (stores the return address)
bffff038 --> ebp (saved base pointer)
bffff02c --> local1 address

Now I must be missing something because there is a 8 byte gap between ebp and local1. I assumed local1 data type is 4 bytes which leaves the address between bffff030 -> bffff038 unaccounted for. Would really appreciate help with this one.

EDIT

here is the assembly code. the only anomaly i could think of is the SUB instruction following mov esp,ebp, though I'm not sure how it relates to the gap. enter image description here

Upvotes: 0

Views: 594

Answers (1)

Valyrie H. Autumn
Valyrie H. Autumn

Reputation: 162

The 8 byte gap is because on x86 calling conventions require 16 byte stack alignment on entrance to a function, and a return pointer is only 8 bytes. So we get the 8 bytes of "wasted" space.

Upvotes: 1

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