如何通過引用傳遞數組,如果數據類型是typedef。我正在學習C++,我閱讀了引用調用的概念,但是當我按照這種方式實現時 - 我得到一個錯誤(在代碼之後粘貼下面)。請任何人都可以解釋發送一個數組作爲參考調用的最佳方式嗎?如何在函數參數中引用數組?
#include <iostream>
#include <vector>
using namespace std;
typedef unsigned long ulong;
ulong fib_dynamic(ulong n, ulong &memo[]){
if(n < 2) return 1;
if(memo[n] == 0){
memo[n] = fib_dynamic(n-1, memo) + fib_dynamic(n-2, memo);
}
return memo[n];
}
ulong fib_iterative(ulong n){
ulong fib[n+1];
fib[0] = 1;
fib[1] = 1;
for(int i=2; i<n; i++) {
fib[i] = fib[i-1] + fib[i-2];
}
return fib[n-1];
}
int main(){
ulong n;
cout << "Welcome to Fib Calculator\nEnter the n:";
cin >> n;
ulong memo[n];
cout << endl << n << " th fib num(dynamic) = " << fib_dynamic(n,memo) << endl;
}
//錯誤
1-fib-dp.cpp:13:47: error: 'memo' declared as array of references of type
'unsigned long &'
ulong fib_dynamic(ulong n, unsigned long &memo[]){
^
1-fib-dp.cpp:37:53: error: no matching function for call to 'fib_dynamic'
cout << endl << n << " th fib num(dynamic) = " << fib_dynamic(n,memo) << endl;
^~~~~~~~~~~
1-fib-dp.cpp:13:7: note: candidate function not viable: no known conversion from
'ulong [n]' to 'int' for 2nd argument
ulong fib_dynamic(ulong n, unsigned long &memo[]){
^
2 errors generated.
爲什麼你覺得你首先需要一個參考?使它成爲'ulong memo []' –
還要注意''ulong memo [n]'有'n'個元素,索引爲'0'到'n-1'。因此'備忘錄[n]'通過緩衝區溢出展現了未定義的行爲。 –
'ulong&memo []'是一個數組(指針)到ulong's('ulong&')的引用。 'ulong(&memo)[]'是對'ulong'數組的引用。 –