哪个递归函数会花费更少的内存？

Question

void func(int depth){
    if(depth== 0) return;
    int number(12345);
    cout << number; //it does something with number
    func(--depth);
}
void func2(int depth){
    if(depth== 0) return;
    {
        int number(12345);
        cout << number; //it does something with number
    }
    func2(--depth);
}
void main(){
    func(10); //will this function cost more memory?
    func2(10);//will this function cost less memory?
}

你好.我这里有两个功能.func2会花费更少的内存,因为它的数字(12345)被"{}"封装,所以当func2调用下一次迭代时,数字(12345)可以超出范围并消失吗？

我相信func会花费更多,因为它的数量(12345)即使到达下一次迭代时也不在范围之外？

Answer 1

假设我们有AMD/Intel x86_64架构,我们的编译器是GCC.

让我们获取装配输出(-O2 -S)并分析:

func:
.LFB1560:
    pushq   %rsi
    .seh_pushreg    %rsi
    pushq   %rbx
    .seh_pushreg    %rbx
    subq    $40, %rsp
    .seh_stackalloc 40
    .seh_endprologue
    movl    %ecx, %ebx
    testl   %ecx, %ecx
    je  .L3
    movq    .refptr._ZSt4cout(%rip), %rsi
    .p2align 4,,10
.L5:
    movl    $12345, %edx
    movq    %rsi, %rcx
    call    _ZNSolsEi
    subl    $1, %ebx
    jne .L5
.L3:
    addq    $40, %rsp
    popq    %rbx
    popq    %rsi
    ret
    .seh_endproc

func2:
.LFB2046:
    pushq   %rsi
    .seh_pushreg    %rsi
    pushq   %rbx
    .seh_pushreg    %rbx
    subq    $40, %rsp
    .seh_stackalloc 40
    .seh_endprologue
    movl    %ecx, %ebx
    testl   %ecx, %ecx
    je  .L10
    movq    .refptr._ZSt4cout(%rip), %rsi
    .p2align 4,,10
.L12:
    movl    $12345, %edx
    movq    %rsi, %rcx
    call    _ZNSolsEi
    subl    $1, %ebx
    jne .L12
.L10:
    addq    $40, %rsp
    popq    %rbx
    popq    %rsi
    ret
    .seh_endproc

如您所见,两个功能完全相同.