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new And delete

When I was a trainee at the training institution. I was shocked when my teacher shows me the secret blow ‘new’ and ‘delete’. I still remember that day even I can’t fully understand it. It’s time to resolve it by myself!

至今我还记得在培训的时候, 讲师给我演示了 new 和 delete 的内部实现时我的震惊. 虽然当时没有完全理解. 是时候自己看看了.

new class

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class A {
public:
    int i = 0;
    char c = 0;
};

int main() {
    A *pa = new A;
    pa->i = 10;
    printf("%d\n", pa->i);
    delete pa;
}
// -------------------------------
main:
.LFB4945:
        .loc 1 11 0
        .cfi_startproc
        pushq   %rbp
        .cafi_def_cfa_offset 16
        .cfi_offset 6, -16
        movq    %rsp, %rbp
        .cfi_def_cfa_register 6
        pushq   %rbx
        subq    $24, %rsp
        .cfi_offset 3, -24
        .loc 1 12 0
        movl    $8, %edi
        call    _Znwm@PLT
        movq    %rax, %rbx
        movq    %rbx, %rdi
        call    _ZN1AC1Ev
        movq    %rbx, -24(%rbp)
        .loc 1 13 0
        movq    -24(%rbp), %rax
        movl    $10, (%rax)
        .loc 1 14 0
        movq    -24(%rbp), %rax
        movl    (%rax), %eax
        movl    %eax, %esi
        leaq    .LC0(%rip), %rdi
        movl    $0, %eax
        call    printf@PLT
        .loc 1 15 0
        movq    -24(%rbp), %rax
        movl    $8, %esi	// 第二个参数是大小
        movq    %rax, %rdi	// 第一个参数是 this
        call    _ZdlPvm@PLT
        .loc 1 16 0
        movl    $0, %eax
        addq    $24, %rsp
        popq    %rbx
        popq    %rbp
        .cfi_def_cfa 7, 8
        ret
        .cfi_endproc

没有什么好说的, 分配空间, 调用构造. 然后析构.

new an array of classes

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class A {
public:
    int i = 0;
    char c = 0;
};

int main() {
    A *pa = new A[2];
    pa->i = 10;
    printf("%d\n", pa->i);
    delete[] pa;
}
// -------------------------------
main:
.LFB4945:
        .loc 1 11 0
        .cfi_startproc
        pushq   %rbp
        .cfi_def_cfa_offset 16
        .cfi_offset 6, -16
        movq    %rsp, %rbp
        .cfi_def_cfa_register 6
        pushq   %r13
        pushq   %r12
        pushq   %rbx
        subq    $24, %rsp
        .cfi_offset 13, -24
        .cfi_offset 12, -32
        .cfi_offset 3, -40
        .loc 1 12 0
        movl    $16, %edi
        call    _Znam@PLT
        movq    %rax, %r13
        movq    %r13, %rax
        movl    $1, %ebx
        movq    %rax, %r12
.L4:
        .loc 1 12 0 is_stmt 0 discriminator 3
        testq   %rbx, %rbx	// 推出循环的条件~
        js      .L3
        .loc 1 12 0 discriminator 2
        movq    %r12, %rdi
        call    _ZN1AC1Ev
        addq    $8, %r12
        subq    $1, %rbx
        jmp     .L4	// 循环~
.L3:
        .loc 1 12 0
        movq    %r13, -40(%rbp)
        .loc 1 13 0 is_stmt 1
        movq    -40(%rbp), %rax
        movl    $10, (%rax)
        .loc 1 14 0
        movq    -40(%rbp), %rax
        addq    $8, %rax
        movl    $20, (%rax)
                .loc 1 15 0
        movq    -40(%rbp), %rax
        movl    (%rax), %eax
        movl    %eax, %esi
        leaq    .LC0(%rip), %rdi
        movl    $0, %eax
        call    printf@PLT
        .loc 1 16 0
        movq    -40(%rbp), %rax
        addq    $8, %rax
        movl    (%rax), %eax
        movl    %eax, %esi
        leaq    .LC0(%rip), %rdi
        movl    $0, %eax
        call    printf@PLT
        .loc 1 17 0
        cmpq    $0, -40(%rbp)
        je      .L5
        .loc 1 17 0 is_stmt 0 discriminator 1
        movq    -40(%rbp), %rax
        movq    %rax, %rdi
        call    _ZdaPv@PLT	// 删除数组只有 this?
.L5:
        .loc 1 18 0 is_stmt 1
        movl    $0, %eax
        addq    $24, %rsp
        popq    %rbx
        popq    %r12
        popq    %r13
        popq    %rbp
        .cfi_def_cfa 7, 8
        ret

也没有什么特别惊艳的, 得看看 _ZdaPv

_ZdaPv

…

跟吐了… 🤮

不知道有多少层 jnmp, 多少个循环… 或许不是现在🤣…

不过, 还好嘛… 至少知道了如何操作的… 虽然不知道细节.

不过根据推断, 分配的只有 16 字节, 那么意思是应该没有额外的空间用于保存数组大小(虽然有可能在函数内部, 但感觉可能性不大)

emm… 果然我还是不安分的人呢…

我果然还是喜欢做这一类的事情.