有人可以解释为什么当我回来形式的功能,我从tabOfOffsets丢失我的数据。我做了两次相同的事情,只有第二个数组的程序崩溃。 我在函数的最后打印了这个数组的值,并且一切都清晰而正确。也许我在删除某处出错了? 下面是代码。C++分配内存
#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;
}
获得一些[良好的初学者书](http://stackoverflow.com/questions/388242/the-definitive-c-book-guide-and-list),并阅读*参考*和如何通过参数*参考*。 –
不要使用指向数组的指针。这是C风格的编程。使用'std :: vector'并通过引用传递。 – JHBonarius
在了解了参考资料并知道如何解决您的程序后,请考虑您所做的*双重复制。为什么不简单地*分配*指针?像'tabValue = newTabOfValues'?一旦你考虑并实施并测试了这些,将你的程序扔掉并学习如何使用['std :: vector'](http://en.cppreference.com/w/cpp/container/vector)。 –