Modulus in Assembly x64 linux question C++ [duplicate] - c++

Question

This question already has answers here:
Why does GCC use multiplication by a strange number in implementing integer division?
(5 answers)
Divide Signed Integer By 2 compiles to complex assembly output, not just a shift
(1 answer)
Closed 1 year ago.
I have these functions in C++
int f1(int a)
{
int x = a / 2;
}
int f2(int a)
{
int y = a % 2;
}
int f3(int a)
{
int z = a % 7;
}
int f4(int a,int b)
{
int xy = a % b;
}
And i saw their assembly code but couldn't understand what they are doing.I couldn't even find a good referance or some explained example for the same. Here is the assembly
f1(int):
push rbp
mov rbp, rsp
mov DWORD PTR [rbp-20], edi
mov eax, DWORD PTR [rbp-20]
mov edx, eax
shr edx, 31
add eax, edx
sar eax
mov DWORD PTR [rbp-4], eax
nop
pop rbp
ret
f2(int):
push rbp
mov rbp, rsp
mov DWORD PTR [rbp-20], edi
mov eax, DWORD PTR [rbp-20]
cdq
shr edx, 31
add eax, edx
and eax, 1
sub eax, edx
mov DWORD PTR [rbp-4], eax
nop
pop rbp
ret
f3(int):
push rbp
mov rbp, rsp
mov DWORD PTR [rbp-20], edi
mov eax, DWORD PTR [rbp-20]
movsx rdx, eax
imul rdx, rdx, -1840700269
shr rdx, 32
add edx, eax
sar edx, 2
mov esi, eax
sar esi, 31
mov ecx, edx
sub ecx, esi
mov edx, ecx
sal edx, 3
sub edx, ecx
sub eax, edx
mov DWORD PTR [rbp-4], eax
nop
pop rbp
ret
f4(int, int):
push rbp
mov rbp, rsp
mov DWORD PTR [rbp-20], edi
mov DWORD PTR [rbp-24], esi
mov eax, DWORD PTR [rbp-20]
cdq
idiv DWORD PTR [rbp-24]
mov DWORD PTR [rbp-4], edx
nop
pop rbp
ret
Can you please tell by some example or what steps it is following to calculate the answers in all these three cases and why would they work just fine instead of normal divide