How can a function be called without the call instruction in assembly? [duplicate]

I'm learning to read assembly code. For that, I wrote this C program (named vebytesmain.c) with level 0 optimization:

typedef char *ptr;

char ret_byte(void *ender, int byte){
    ptr point = (ptr)(ender + byte);
    return *point;
}

int main(void){
    int num=0x00010203;
    num = ret_byte(&num,2);
    return 0;
}

and got it's assembly code by gcc -O0 -S vebytesmain.c. The main function was like that:

main:
.LFB1:
    .cfi_startproc
    pushq   %rbp
    .cfi_def_cfa_offset 16
    .cfi_offset 6, -16
    movq    %rsp, %rbp
    .cfi_def_cfa_register 6
    subq    $16, %rsp
    movq    %fs:40, %rax
    movq    %rax, -8(%rbp)
    xorl    %eax, %eax
    movl    $66051, -12(%rbp)
    leaq    -12(%rbp), %rax
    movl    $2, %esi
    movq    %rax, %rdi
    call    ret_byte        ;    this is where the call happens (lvl 0 optimization)
    movsbl  %al, %eax
    movl    %eax, -12(%rbp)
    movl    $0, %eax
    movq    -8(%rbp), %rdx
    xorq    %fs:40, %rdx
    je  .L5
    call    __stack_chk_fail@PLT
.L5:
    leave
    .cfi_def_cfa 7, 8
    ret
    .cfi_endproc
.LFE1:
    .size   main, .-main
    .ident  "GCC: (Ubuntu 7.4.0-1ubuntu1~18.04.1) 7.4.0"
    .section    .note.GNU-stack,"",@progbits

which is ok, but when I apply at least level 2 optimization, what I get is:

main:
.LFB1:
    .cfi_startproc
    xorl    %eax, %eax
    ret
    .cfi_endproc
.LFE1:
    .size   main, .-main
    .ident  "GCC: (Ubuntu 7.4.0-1ubuntu1~18.04.1) 7.4.0"
    .section    .note.GNU-stack,"",@progbits

How come there is no call instruction for the ret_bytes function? How do I know that the function is being called? How is it being called?

I would be grateful if someone could solve explain me what is going on.