十進制到十六進制程序代理

我目前正在研究將十進制輸入轉換爲十六進制等效的程序。我計劃使用矢量來系統地收集這些值，然後將它們反向吐出（因此以十六進制表示）。但是我遇到了很多問題。所有的投入都會按照預期工作，然後事情會變得很奇怪。輸入16將導致系統輸出一個笑臉，在2個不同的彩色笑臉中輸入18個結果，23個導致系統發出嘟嘟聲。十進制到十六進制程序代理

這應該是一個相對直接的代碼，它只是以我從未見過的方式行事！希望這個信息可以幫助

編輯：我知道std :: hex函數，雖然這是一個類，我們不允許使用它unfortunatley。

#include <iostream> 
#include <string> 
#include <vector> 
#include <iomanip> 

using namespace std; 

int main(){ 

int input, tester = 0, switchStuff, county = 0, remainderCount = 1, inputCount, remainder = 1, remainderCount2, inputCount2; 
string stopGo; 
char remainderLetter; 

do{ 

    while (tester != 1){ 
    cout << "Enter the number you would like to convert to Hex format: "; 
    cin >> input; 
    if (cin.fail()){      //check if user input is valid 
    cout << "Error: that is not a valid integer.\n"; 
    cin.clear(); cin.ignore(INT_MAX, '\n');  //Clear input buffer 
    continue; //continue skips to top of loop if user input is invalid, allowing another attempt 
    } 
    else{ 
    tester = 1;  //Tester variable allows loop to end when good value input 
    } 
} 

inputCount = input; 

while(inputCount != 0){ 

    remainderCount = inputCount % 16; 
    inputCount = (inputCount - remainderCount)/16; 
    county++; 

    } 

vector<string>userInfo(county); 

inputCount2 = input; 

for (int i = 0; county > i; i++){ 

    remainderCount2 = inputCount2 % 16; 
    inputCount2 = (inputCount2 - remainderCount2)/16; 

    if (remainderCount2 == 10){ 
    remainderLetter = 'A'; 
    } 
    else if (remainderCount2 == 11){ 
    remainderLetter = 'B'; 
    } 
    if (remainderCount2 == 12){ 
    remainderLetter = 'C'; 
    } 
    else if (remainderCount2 == 13){ 
    remainderLetter = 'D'; 
    } 
    if (remainderCount2 == 14){ 
    remainderLetter = 'E'; 
    } 
    else if (remainderCount2 == 15){ 
    remainderLetter = 'F'; 
    } 

    if (remainderCount2 >= 10){ 
    userInfo[i] = remainderLetter; 
    } 
    else{ 
    userInfo[i] = remainderCount2; 
    } 


} 

cout << "The result in Hexadecimal format is: ";  

for (int i = 0; i < county; i++) 
    cout << userInfo[i]; 

cout << endl << "Would you like to continue? (Enter Yes/No): ";  //Check whether to continue or not 
cin >> stopGo; 
cout << endl; 

tester = 0; 

}while ((stopGo.compare("Yes") == 0) || (stopGo.compare("yes") == 0) || (stopGo.compare("y") == 0) || (stopGo.compare("Y") == 0)); //Leaves user with a range of ways to say 'yes' 

cout << "Thank you for using this program!" << endl; 

system("pause"); 
}

來源

2014-12-04 Dozar

您是否知道，您可以使用std :: hex在std :: out上以十六進制格式打印任何整數數字？ – MikeMB 2014-12-04 08:01:03

也許你會考慮切換到這樣的： cout << hex << numberToConvert; numberToConvert - 你想要轉換爲十進制格式的數字它會做同樣的事情，你只需要改變輸出數據格式。 – MateuszZawadzki 2014-12-04 08:01:56

沒有時間給出完整的答案，但代碼的最後一部分很糟糕：考慮到'A'的ASCII代碼是65，你可以使用類似'remainingLetter = remainderCount2 + 55;'或者更好的'remainingLetter = remainderCount2 +'A'-10;' – kebs 2014-12-04 08:17:52

嘗試改變

userInfo[i] = remainderCount2;

到

userInfo[i] = remainderCount2 + '0';

（ '原因數字0,1，...是不一樣的字符 '0'，' 1' ， ...）

順便說一句，有一個簡單的方法來十進制轉換爲十六進制

std::string sDecimal[] = "155"; 
char xHex[50]; 
sprintf(xHex, "%X", atoi(sDecimal.c_str()));

來源

2014-12-04 08:11:39 Fl0

十進制到十六進制程序代理

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