Question

有人可以解释一下为什么当我退回表单功能时我从tabOfOffsets丢失了我的数据。我做了两次相同的事情，只用第二个数组编程崩溃。我在函数末尾打印了这个数组的值，一切都清晰正确。也许我在某处删除错误？下面是代码。

#include<iostream>
#include <algorithm>

using std::cout;
using std::endl;

void changeSizeOfVector(int *tabValue, int *tabOffsets, int &oldSize, int 
newSize) {
int temp = std::min(oldSize, newSize);

int *newTabOfValues = new int [newSize] {0};
int *newTabOfOffsets = new int [newSize] {0};

for (int i = 0; i < temp; i++) {
    newTabOfValues[i] = tabValue[i];
    newTabOfOffsets[i] = tabOffsets[i];
}

delete[] tabValue;
delete[] tabOffsets;

tabValue = new int [newSize] {0};
tabOffsets = new int [newSize] {0};



for (int i = 0; i < newSize; i++) {
    tabValue[i] = newTabOfValues[i];
    tabOffsets[i] = newTabOfOffsets[i];
    std::cout << tabOffsets[i] << tabValue[i] << endl;
}

oldSize = newSize;
delete[] newTabOfValues;
delete[] newTabOfOffsets;
for (int i = 0; i < newSize; i++) {
    std::cout << tabOffsets[i] << tabValue[i] << endl;
}

}
int main() {

int SIZE = 10;

int * tabOfOffsets = new int[SIZE];
int * tabOfValues = new int[SIZE];

for (int i = 0; i < SIZE; i++)
{
    tabOfValues[i] = i;
    tabOfOffsets[i] = i;
    cout << tabOfValues[i] << " : " << tabOfOffsets[i] << endl;
}

changeSizeOfVector(tabOfValues, tabOfOffsets, SIZE, 12);


for (int i = 0; i < SIZE; i++) {
    cout << tabOfOffsets[i] << " : " << tabOfValues[i] << endl;

}
delete[] tabOfOffsets;
delete[] tabOfValues;

}

Answer 1

此函数声明错误：

void changeSizeOfVector(int *tabValue, int *tabOffsets, int &oldSize, int 
newSize);

这意味着您可以更改tabOffsets的值而不是指针本身的值，以使其行为正确您应该如下声明：

void changeSizeOfVector(int *tabValue, int **tabOffsets, int &oldSize, int 
newSize);

通过这种方式，您可以更改指针本身并为其分配新分配的数组。

C ++分配内存

1 个答案: