进入新功能时rsp不动

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When entering in a C function I expected to see in the disassembly how the stack pointer gets subtracted enough to make space for variables, but no; I only see how the address of variables is directly access through ebp, when esp still points to ebp.

push %rbp mov %rsp,%rbp movl $0x4,-0x4(%rbp) mov $0x0,%eax pop %rbp retq

I had to create a lot of variables and initialize them to be taken serious by the computer and see how a lot of unneeded space was made. Was the difference really the amount of space used or something else? and if so, how is it that making room by moving rsp is only needed when I request for a lot of space?

解决方案

The x86-64 System V ABI has a 128-byte red zone below RSP that is safe from asynchronous clobbering, i.e. is "owned" by the function.

It looks like you compiled int foo{ int x = 4; return 0; } with gcc -O0 (optimizations disabled), and gcc chose to keep x in the red-zone instead of adjusting rsp to "reserve" / "allocate" the stack space. (See the red-zone tag wiki for more links / info.)

This is the whole point of the red-zone: to save those sub/add instructions in leaf functions.

BTW, looking at un-optimized code is usually a waste of time. -O1 at least is more readable, and -O2 / -O3 are relevant for code that you should actually care about. See also How to remove "noise" from GCC/clang assembly output?.

In programs with no signal handlers, the entire stack region can effectively be used as a red-zone. example: code-golf extended-precision Fibonacci using esp as an array pointer because pop is fast and compact. (AFAIK, signal handlers are the only thing that will asynchronously clobber memory below rsp). The red-zone lets compilers take advantage of it without needing a special compile option (and there is no such option for 32-bit mode where the SysV ABI doesn't define a red-zone). Proving that there are no signal handlers is probably not feasible even with whole-program optimization.

I only see how the address of variables is directly access through ebp

No, you don't. Access through ebp would fault in 64-bit code, because the stack is outside the low 4GB of address space (by default on Linux at least). Pointers are 64-bit, so gcc uses rbp to access them.

Using an address-size prefix to encode movl $0x4,-0x4(%ebp) in 64-bit mode would be a waste of code size even if it didn't fault.

Fun fact: in the x32 ABI (ILP32 in long mode) where pointers are 32-bit, gcc often uses the address-size prefix instead of extra instructions to truncate possible high garbage in registers and make sure addressing modes wrap at 2^32 instead of going outside 4GB (with a signed displacement for example). Instead of doing this optimally, it defaults to just stupidly using address-size prefixes on every instruction with an explicit memory operand. But a recent patch gets it to always use 64-bit rsp instead of using address-size prefixes even for esp.