关于析构函数：C 堆栈分配的变量未销毁(/destroyed?)

C++ stack allocated variable not destructed (/destroyed?)

我对 C 语言很陌生，但我认为我说得对，在堆栈上声明的对象应该在超出范围时自动销毁/销毁？在我目前正在使用的迷你项目中，情况并非如此。

1
2
3
4
5
6
7
8
9

void MainWindow::clickTest() {
FunkyNumber num = 4;
FunkyNumber num2 = 6;

num += num2;
std::cout << num << std::endl; // This works okay!

// Should be destroyed here!
}

我的析构函数应该这样做：

1
2
3
4

virtual FunkyNumber::~FunkyNumber() {
std::cout <<"goodbye cruel world! (" << m_intValue <<")" << std::endl;
// m_intValue is just the int value of this"FunkyNumber"
}

但是标准输出没有任何结果！

可能应该提到我正在使用 Qt - 但这只是一个普通的 C 类，所以从我能看出的情况来看这并不重要...

编辑：
funkynumber.cpp:

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32

#include"funkynumber.h"

FunkyNumber::FunkyNumber(int num)
: m_intValue(num) {
std::cout <<"made a funkynumber" << num << std::endl;
}

FunkyNumber::~FunkyNumber() {
std::cout <<"goodbye cruel world! (" << m_intValue <<")" << std::endl;
}

int FunkyNumber::intValue() const {
return m_intValue;
}

void FunkyNumber::operator+=(const FunkyNumber &other) {
m_intValue += other.intValue();
}

void FunkyNumber::operator=(const FunkyNumber &other) {
m_intValue = other.intValue();
}

bool FunkyNumber::operator==(const FunkyNumber &other) {
return other.intValue() == m_intValue;
}

std::ostream &operator<<(std::ostream &outStream, const FunkyNumber &num) {
outStream <<"FunkyNumber (" << num.intValue() <<")";

return outStream;
}

相关讨论

我无法重现该行为。

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50

#include<iostream>

struct FunkyNumber{
int m_intValue;
FunkyNumber::FunkyNumber(int num)
: m_intValue(num) {
std::cout <<"made a funkynumber" << num << std::endl;
}

FunkyNumber::~FunkyNumber() {
std::cout <<"goodbye cruel world! (" << m_intValue <<")" << std::endl;
}

int FunkyNumber::intValue() const {
return m_intValue;
}

void FunkyNumber::operator+=(const FunkyNumber &other) {
m_intValue += other.intValue();
}

void FunkyNumber::operator=(const FunkyNumber &other) {
m_intValue = other.intValue();
}

bool FunkyNumber::operator==(const FunkyNumber &other) {
return other.intValue() == m_intValue;
}
};

std::ostream &operator<<(std::ostream &outStream, const FunkyNumber &num) {
outStream <<"FunkyNumber (" << num.intValue() <<")";

return outStream;
}

void call(){
FunkyNumber num = 4;
FunkyNumber num2 = 6;

num += num2;
std::cout << num << std::endl; // This works okay!

// Should be destroyed here!
}

int main(int argc, char **argv){
call();
std::cout <<"call ended" << std::endl;
}

这很好用。人们推广 SSCCE 的原因不仅是为了更容易帮助您，还因为它可以帮助您自己找到问题所在(这显然不在您发布的代码中)。