我在用斐波那契数列打印矩阵时遇到麻烦。当我启动程序时,它将打印一个带有整数和其余零的序列。我希望得出一个结论。
顺序打印功能:
void printArray(int _A[], int _sz)
{
_A = new int(_sz);
cout << ">>> [ ";
for (int i = 0; i <= _sz; i++)
{
cout << _A[i] << " ";
}
cout << " ]\n";
}
计算序列的功能:
int fib(int _limit, int *_A)
{
int count = 0;
int fib0 = 1;
int fib1 = 1;
int fib2 = 0;
while (fib2 < _limit)
{
fib2 = fib0 + fib1;
count++;
fib0 = fib1;
fib1 = fib2;
}
_A = new int(count);
for (int i = 0; i < count; i++)
{
cout << _A[i] << " ";
}
return count;
}
主要:
int main()
{
int L;
int *A = nullptr;
cout << ">>> Press number: "
cin >> L;
if (L >= 0)
{
auto sz = fib(L, A);
cout << ">>> The Fibonacci series up to " << L << " is: \n";
printArray(A, sz); .
}
else
cout << ">>> Sorry, only positive integers accepted. Aborting...\n";
}
您有很多问题。
在printArray
中,您用单个整数指针_A = new int(_sz);
覆盖了传入的数组,尝试打印数组的所有元素将是未定义的行为。
在fib
中,您还使用了new int(count);
,这再次分配了一个指针而不是数组。您应该改用new int[count];
。您永远不会将值分配给数组_A
的元素,因此它们不会被初始化,并且使用它们是不确定的行为。
您不会将数组_A
传递回调用函数。您可以通过将fib
的声明更改为:
int fib(int _limit, int *&_A)
解决所有这些问题的一种好方法是使用std::vector
而不是数组,那么您无需预先计算序列的大小:
#include <iostream>
#include <vector>
using std::cout;
using std::cin;
using std::vector;
void printArray(const std::vector<int>& A)
{
cout << ">>> [ ";
for (int i : A)
{
cout << i << " ";
}
cout << " ]\n";
}
void fib(int _limit, std::vector<int>& A)
{
int fib0 = 1;
A.push_back(fib0);
int fib1 = 1;
int fib2 = 0;
while (fib2 < _limit)
{
A.push_back(fib1);
fib2 = fib0 + fib1;
fib0 = fib1;
fib1 = fib2;
}
}
int main()
{
int L;
std::vector<int> A;
cout << ">>> Press number: ";
cin >> L;
if (L >= 0)
{
fib(L, A);
cout << ">>> The Fibonacci series up to " << L << " is: \n";
printArray(A);
}
else
cout << ">>> Sorry, only positive integers accepted. Aborting...\n";
}
[如果您正在学习一些反常的计算机科学课程,这些课程在教您c ++,但不允许您使用c ++类,那么使用数组的相同代码可能看起来像:
#include <iostream>
using std::cout;
using std::cin;
void printArray(int A[], int _sz)
{
cout << ">>> [ ";
for (int i = 0; i <= _sz; i++)
{
cout << A[i] << " ";
}
cout << " ]\n";
}
int fib(int _limit, int *A)
{
int count = 0;
int fib0 = 1;
if (A) A[0] = fib0;
int fib1 = 1;
int fib2 = 0;
while (fib2 < _limit)
{
count++;
if (A) A[count] = fib1;
fib2 = fib0 + fib1;
fib0 = fib1;
fib1 = fib2;
}
return count;
}
int main()
{
int L;
int *A = nullptr;
cout << ">>> Press number: ";
cin >> L;
if (L >= 0)
{
auto sz = fib(L, A);
A = new int[sz];
fib(L, A);
cout << ">>> The Fibonacci series up to " << L << " is: \n";
printArray(A, sz);
}
else
cout << ">>> Sorry, only positive integers accepted. Aborting...\n";
}