# **資料型態/轉型** ### 2020 資訊之芽 #### 2020/3/28 #### 實習生 袁紹奇 --- ## Outline * 資料型態(float, double) * 控制輸出 * 轉型 --- ## 資料型態?? ---- ### 資料型態 * 變數的儲存方式 * 不同型態的變數所占大小不同 * bit? byte? * 8 bits = 1 byte ---- | 型別 | 意義 | | --------- | --------------- | | bool | boolean值(真值) | | char | 字元 | | int | 整數 | | short | 短整數 | | long | 長整數 | | long long | 更長的整數 | | float | 浮點數 | | double | 倍準浮點數 | --- ### 布林 bool * true OR false * 1 和 0 * [George Boole](https://zh.wikipedia.org/wiki/%E4%B9%94%E6%B2%BB%C2%B7%E5%B8%83%E5%B0%94) * 許多比較運算式的回傳值 * 其實沒有比int小 * 非0就是true ---- ## 一個小例子 ```cpp bool a = 9527; if(a) cout << "yes"; ``` --- ## 可是，電腦只能存0和1阿 ## >____< ---- ## int是怎麼存的? ---- ## int是怎麼存的? 世界上有10種人，一種是懂二進位的，另一種不懂 ---- ### 整數 int * integer * 二進位儲存 * 有號(signed) * 範圍?!?!?! ----  ---- ### 整數 int * 通常以4個bytes儲存 * sizeof(int) * 範圍是$[−2^{31}$ ~ $2^{31}−1]$ * $2^{31}$ = 2147483648 * 無號(unsigned)? ---- ## OVERFLOW!!! ```cpp int a = 2147483647; cout << a << endl; // 2147483647 a++; cout << a << endl; // -2147483648 ``` ---- ## wraps around? ```cpp unsigned int a = 6666; int i = -9527; cout << i + i << endl; // -19054 cout << i + a << endl; // 4294964435 ``` ----  --- ## 分數怎麼存? ---- ## Recall ```cpp cout << 5/2 << endl; // 會輸出2 ``` ---- ### float/double * 浮動的點點 * 4 bytes/ 8 bytes * 沒有unsigned double?!  ---- ### 一些奇怪的double * NAN * INF * e+/e- ---- ## NAN ```cpp double a = sqrt(-1.0); // #include <cmath> for sqrt cout << a << endl; // output: nan (not a number) ``` ---- ## INF ```cpp double a = 2.0/0.0; cout << a << endl; // output: inf (infinite) ``` ---- ## e+/e- ```cpp cout << 123456789.1234 << endl; // 1.23457e+008 cout << 0.00000000000000123 << endl; // 1.23e-015 //////// ***浮動的點點*** //////// ``` ---- ## 練習(演示) 練習輸入一個整數n，輸出以此為半徑的圓面積 pi = 3.14159 --- ## 每次輸出的小數點位數不一樣? ---- ## iomanip * #include <iomanip> * cout 預設為整數加小數點後共六位(整數部分優先) * cout.precision() (設定位數) * fixed (指定小數點後位數) * setprecision() (設定位數) ---- ```cpp double a = 123.4567899527; cout << a << endl; // 123.457 // 四捨五入? cout << fixed << setprecision(8) << a << endl; // 123.45678995 // fixed 會讓之後的precision變成設定小數點後位數 cout.precision(10); cout << a; // 123.4567899527 ``` --- 練習:弧度量轉度度量  度度量 = 弧度量 * 360/2pi 假設 pi = 3.14159 輸出到小數點後四位 ---- ## solution ```cpp #include <iostream> #include <iomanip> using namespace std; int main(){ double pi = 3.14159; int a; double ans = 0.0; cin >> a; ans = a * 360 / 2 / pi; cout << fixed << setprecision(4) << ans; } ``` ---- ## 等等... ```cpp ans = a * 360 / 2 / pi; ^ // 這邊有個int? ``` --- ## 型態轉換 * 自己手動轉 * 編譯器幫你轉 * 不會改變變數型別 ```cpp // compile error int wuhan[9527]; int feyan = wuhan[sqrt(9)*5+27]; // #include <math.h> for sqrt() ``` ---- ## 兩種轉型(手動轉) ```cpp cout << (int)3.14159 << endl; // 3 cout << (float)5 / (float)2 << endl; // 2.5 cout << (float)(5/2) << endl; // 2 cout << (char)(97) << endl; // a cout << static_cast <int>(3.14159) << endl; cout << static_cast <float>(5) / static_cast <float>(2) << endl; cout << static_cast <float>(5/2) << endl; cout << static_cast <char>(97) << endl; ``` ---- ## 編譯器大哥幫忙 * 整數與浮點數相加 * 小數與大數相加 * 有號與無號相加 ```cpp cout << 3.14 + 1; // 4.14 cout << (int)(3.14) + (double)(1.2); // 4.2 cout << 5 / 2 + 0.5; // 2.5 cout << 5 / 2.0 + 0.5; // 3.0 cout << (unsigned int)INT_MAX + (int) 1; // 2147483648 ``` ---- ## 型態轉換 * 位階小的和位階大的運算會被轉成大的 * bool < char < short < int < long < long long * 有小數先轉小數 * float < double < long double * 同位階的unsigned 比原本的大 --- ## 練習 ```cpp cout << (unsigned int)1 + (double)-2.2 << endl; cout << (unsigned)2147483647 + 1.0 << endl; cout << (unsigned)10 - 20 << endl; cout << (long long)-1 + (unsigned)0 << endl; ``` ---- ## 練習 ```cpp cout << (unsigned int)1 + (double)-2.2 << endl; // -1.2 cout << (unsigned)2147483647 + 1.0 << endl; cout << (unsigned)10 - 20 << endl; cout << (long long)-1 + (unsigned)0 << endl; ``` ---- ## 練習 ```cpp cout << (unsigned int)1 + (double)-2.2 << endl; // -1.2 cout << (unsigned)2147483647 + 1.0 << endl; // 2.14748e+009 cout << (unsigned)10 - 20 << endl; cout << (long long)-1 + (unsigned)0 << endl; ``` ---- ## 練習 ```cpp cout << (unsigned int)1 + (double)-2.2 << endl; // -1.2 cout << (unsigned)2147483647 + 1.0 << endl; // 2.14748e+009 cout << (unsigned)10 - 20 << endl; // 4294967286 cout << (long long)-1 + (unsigned)0 << endl; ``` ---- ## 練習 ```cpp cout << (unsigned int)1 + (double)-2.2 << endl; // -1.2 cout << (unsigned)2147483647 + 1.0 << endl; // 2.14748e+009 cout << (unsigned)10 - 20 << endl; // 4294967286 cout << (long long)-1 + (unsigned)0 << endl; //-1 ``` ---