3 ms·

This is because of the calling convention / ABI If you write void free_with_extra_args(int a, int b) { free(a); } then *a is in already in

This is because of the calling convention / ABI If you write void free_with_extra_args(int *a, int *b) { free(a); } then *a is in already in the correct slot (the RDI register) for the first argument when free_with_extra_args is being called. Whatever is put into *b is never touched. If you compile this with gcc -O2 you get free_with_extra_args: jmp free If you make the function call free(b) instead, you'll have to move b into the right place before calling free: free_with_extra_args: mov rdi, rsi jmp free This is on x86-64 as summarized here https://en.wikipedia.org/wiki/X86_calling_conventions#System_V_AMD64_ABI https://en.wikipedia.org/wiki/X86_calling_conventions#System... Wikipedia also has a nice summary of calling conventions on other platforms like ARM. All modern calling conventions are similar: pass the first args in registers and then use the stack as needed https://en.wikipedia.org/wiki/Calling_convention https://en.wikipedia.org/wiki/Calling_convention

flohofwoe 3y ago
That's only one half of it though, the other interesting part is the jmp (instead of call) to hand over control to a subroutine without pushing a new return address to the stack (since there's no code after the function call, and the calling function doesn't require its own stack frame).
- jonathrg 3y ago
  Indeed. Typically you have pairs of call and ret, where call creates a stack frame and ret tears it down caller caller | ^ (call) . . . . . . . . . . . . . (ret) | | V | outer_function outer_function | ^ (call) . . . .(ret) | | V | inner_function A jmp does not modify the stack, so when the inner function calls ret it jumps right back to caller caller caller | ^ (call) . . . . . . (ret) | | v | outer_function | | | (jmp) | | | V | inner_function This trick stops working as soon as outer_function needs local variables or does anything other than returning the exact return value of inner_function. In that case you need a stack frame

This is because of the calling convention / ABI If you write void free_with_extra_args(int *a, int *b) { free(a); } then *a is in already in

flohofwoe 3y ago

That's only one half of it though, the other interesting part is the jmp (instead of call) to hand over control to a subroutine without pushing a new return address to the stack (since there's no code after the function call, and the calling function doesn't require its own stack frame).

jonathrg 3y ago
Indeed. Typically you have pairs of call and ret, where call creates a stack frame and ret tears it down caller caller | ^ (call) . . . . . . . . . . . . . (ret) | | V | outer_function outer_function | ^ (call) . . . .(ret) | | V | inner_function A jmp does not modify the stack, so when the inner function calls ret it jumps right back to caller caller caller | ^ (call) . . . . . . (ret) | | v | outer_function | | | (jmp) | | | V | inner_function This trick stops working as soon as outer_function needs local variables or does anything other than returning the exact return value of inner_function. In that case you need a stack frame