這篇是想放一些神奇的C++語法錯誤
也可能會放基礎的(X
然後可能會是動態更新
lambda capture
(Update: 2021/1/21)
這似乎是因為capture到值的時候還沒成功建構func這個變數，所以會出問題
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| #include <iostream>
#include <functional>
using namespace std;
signed main() {
function<void(void)> func = [=]() {
int x;
cin >> x;
cout << "ok " << x << '\n';
if(x) func();
};
func();
}
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const reference & implicit conversion
(Update: 2021/1/21)
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| #include <iostream>
using namespace std;
struct Data {
int x;
Data(int val = 0) : x(val) {}
int calc() {
return x * 2 + 3;
}
};
istream & operator>>(istream &I, const Data &data) {
return I >> data.x;
}
int main() {
Data data;
cin >> data;
cout << data.calc() << '\n';
}
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vector<bool> access with auto deduce type
(Update: 2021/1/21)
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| #include <iostream>
#include <vector>
using namespace std;
int main() {
const int n = 20, m = 20;
// n * m grid
auto check = [&](int x, int y) {
cout << x*m+y << '\n';
vector<bool> ok(n * m);
// do some calc
return ok[x * m + y];
};
cout << check(0, 0) << '\n';
}
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abs outside of std
(Update: 2021/8/15)
std::abs 有多型，會自己偵測吃進去的是不是 long long，但是 std 外面的 abs 不會，要用 llabs。
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| #include <bits/stdc++.h>
using ll = long long;
signed main() {
ll n = 5298309920314;
std::cout << abs(n) << std::endl; // overflow!
std::cout << llabs(n) << std::endl;
std::cout << std::abs(n) << std::endl;
}
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Array in Temporary Object
(Update: 2021/9/19)
如果要在 struct 裡面使用指標或是原生陣列的話，要注意如果你保留一個臨時物件的這種成員，可能會讓這個指標在臨時物件被銷毀之後變得指向一塊不合法的記憶體。
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| #include <iostream>
#include <map>
using namespace std;
struct F {
int a[2];
F() : a{0, 1} {}
};
void print(int a[2]) {
for (int i = 0; i < 2; i++)
cerr << a[i];
cerr << '\n';
}
map<int,F> mp;
F GET(int i) {
return mp[i];
}
void safe() {
cerr << "--- safe BEGIN ---\n";
auto B = GET(0);
auto a = B.a;
print(a);
cerr << "--- safe END -----\n";
}
void notsafe() {
cerr << "--- notsafe BEGIN ---\n";
auto a = GET(0).a;
print(a);
cerr << "--- notsafe END -----\n";
}
signed main() {
F A;
mp[0] = A;
print(A.a);
safe();
notsafe();
}
|