Why does the compiler allocate more than needed in the stack? - c

Question

I have a simple C program. Let's say, for example, I have an int and a char array of length 20. I need 24 bytes in total.
int main()
{
char buffer[20];
int x = 0;
buffer[0] = 'a';
buffer[19] = 'a';
}
The stack needs to be aligned to a 16 bytes boundary, so I presume a compiler will reserve 32 bytes. But when I compile such a program with gcc x86-64 and read the output assembly, the compiler reserves 64 bytes.
..\gcc -S -o main.s main.c
Gives me:
.file "main.c"
.def __main; .scl 2; .type 32; .endef
.text
.globl main
.def main; .scl 2; .type 32; .endef
.seh_proc main
main:
pushq %rbp # RBP is pushed, so no need to reserve more for it
.seh_pushreg %rbp
movq %rsp, %rbp
.seh_setframe %rbp, 0
subq $64, %rsp # Reserving the 64 bytes
.seh_stackalloc 64
.seh_endprologue
call __main
movl $0, -4(%rbp) # Using the first 4 bytes to store the int
movb $97, -32(%rbp) # Using from RBP-32
movb $97, -13(%rbp) # to RBP-13 to store the char array
movl $0, %eax
addq $64, %rsp # Restoring the stack with the last 32 bytes unused
popq %rbp
ret
.seh_endproc
.ident "GCC: (x86_64-posix-seh-rev0, Built by MinGW-W64 project) 5.2.0"
Why is that? When I program assembly, I always reserve only the minimum memory I need without any problem. Is that a limitation of the compiler which has trouble evaluating the needed memory or is there a reason for that?
Here is gcc -v
Using built-in specs.
COLLECT_GCC=gcc
COLLECT_LTO_WRAPPER=D:/Mingw64/bin/../libexec/gcc/x86_64-w64-mingw32/5.2.0/lto-wrapper.exe
Target: x86_64-w64-mingw32
Configured with: ../../../src/gcc-5.2.0/configure --host=x86_64-w64-mingw32 --build=x86_64-w64-mingw32 --target=x86_64-w64-mingw32 --prefix=/mingw64 --with-sysroot=/c/mingw520/x86_64-520-posix-seh-rt_v4-rev0/mingw64 --with-gxx-include-dir=/mingw64/x86_64-w64-mingw32/include/c++ --enable-shared --enable-static --disable-multilib --enable-languages=c,c++,fortran,objc,obj-c++,lto --enable-libstdcxx-time=yes --enable-threads=posix --enable-libgomp --enable-libatomic --enable-lto --enable-graphite --enable-checking=release --enable-fully-dynamic-string --enable-version-specific-runtime-libs --disable-isl-version-check --disable-libstdcxx-pch --disable-libstdcxx-debug --enable-bootstrap --disable-rpath --disable-win32-registry --disable-nls --disable-werror --disable-symvers --with-gnu-as --with-gnu-ld --with-arch=nocona --with-tune=core2 --with-libiconv --with-system-zlib --with-gmp=/c/mingw520/prerequisites/x86_64-w64-mingw32-static --with-mpfr=/c/mingw520/prerequisites/x86_64-w64-mingw32-static --with-mpc=/c/mingw520/prerequisites/x86_64-w64-mingw32-static --with-isl=/c/mingw520/prerequisites/x86_64-w64-mingw32-static --with-pkgversion='x86_64-posix-seh-rev0, Built by MinGW-W64 project' --with-bugurl=http://sourceforge.net/projects/mingw-w64 CFLAGS='-O2 -pipe -I/c/mingw520/x86_64-520-posix-seh-rt_v4-rev0/mingw64/opt/include -I/c/mingw520/prerequisites/x86_64-zlib-static/include -I/c/mingw520/prerequisites/x86_64-w64-mingw32-static/include' CXXFLAGS='-O2 -pipe -I/c/mingw520/x86_64-520-posix-seh-rt_v4-rev0/mingw64/opt/include -I/c/mingw520/prerequisites/x86_64-zlib-static/include -I/c/mingw520/prerequisites/x86_64-w64-mingw32-static/include' CPPFLAGS= LDFLAGS='-pipe -L/c/mingw520/x86_64-520-posix-seh-rt_v4-rev0/mingw64/opt/lib -L/c/mingw520/prerequisites/x86_64-zlib-static/lib -L/c/mingw520/prerequisites/x86_64-w64-mingw32-static/lib '
Thread model: posix
gcc version 5.2.0 (x86_64-posix-seh-rev0, Built by MinGW-W64 project)

Answer 1

Compilers may indeed reserve additional memory for themselves.
Gcc has a flag, -mpreferred-stack-boundary, to set the alignment it will maintain. According to the documentation, the default is 4, which should produce 16-byte alignment, which needed for SSE instructions.
As VermillionAzure noted in a comment, you should provide your gcc version and compile-time options (use gcc -v to show these).

Answer 2

Because you haven't enabled optimization.
Without optimization, the compiler makes no attempt to minimize the amount of space or time it needs for anything in the generated code -- it just generates code in the most straight-forward way possible.
Add -O2 (or even just -O1) or -Os if you want the compiler to produce decent code.

Answer 3

I need 24 bytes in total.
The compiler needs space for a return address and a base pointer. As you are in 64 bit mode, that's another 16 bytes. Total 40. Round that up to a 32-byte boundary and you get 64.