将整数转换为字符

Convert integers to characters

本文关键字:字符 转换 整数      更新时间:2023-10-16

我试图让我的程序打印字母而不是数字。我曾经用char c = static_cast<char>(N);来尝试这样做,但它不起作用,而是打印不是 (a-z) 的字符图像 如何让数字打印为字母?

#include  <cstdlib>
#include  <iostream>
using namespace std;
// Function getUserInput obtains an integer input value from the user.
//  This function performs no error checking of user input.
int getUserInput()
{
    int N(0);
    cout << endl << "Please enter a positive, odd integer value, between (1-51): ";
    cin >> N;
    if (N < 1 || N > 51 || N % 2 == 0)
    {
        cout << "Error value is invalid!" << "n";
        cout << endl << "Please enter a positive, odd integer value, between (1-51): ";
        cin >> N;
        system("cls");
    }
    cout << endl;
    return N;
} // end getUserInput function
//  Function printDiamond prints a diamond comprised of N rows of asterisks.
//  This function assumes that N is a positive, odd integer.
void printHourglass(int N)
{
    char c = static_cast<char>(N);
    for (int row = (N / 2); row >= 1; row--)
    {
        for (int spaceCount = 1; spaceCount <= (N / 2 + 1 - row); spaceCount++)
            cout << ' ';
        for (int column = 1; column <= (2 * row - 1); column++)
            cout << c;
        cout << endl;
    } // end for loop
    // print top ~half of the diamond ...
    for (int row = 1; row <= (N / 2 + 1); row++)
    {
        for (int spaceCount = 1; spaceCount <= (N / 2 + 1 - row); spaceCount++)
            cout << ' ';
        for (int column = 1; column <= (2 * row - 1); column++)
            cout << c;
        cout << endl;
    } // end for loop
    // print bottom ~half of the diamond ...

    return;
} // end printDiamond function
int main()
{
    int N = 1;
    while (N == 1)
    {
        printHourglass(getUserInput());
        cout << endl;
        cout << "Would you like to print another hourglass? ( 1 = Yes, 0 = No ):";
        cin >> N;
    }
} // end main function

字母没有编号,A1 开头或类似的东西。您可能使用的是 ASCII/UTF-8 系统。因此,在 printHourglass 中,将cout << N替换为

cout << static_cast<char>('A' + count - 1);
  1. C 函数,itoa
  2. C++,使用字符串流
  3. 提升::lexical_cast

实际上对于您的情况,您可以直接打印出来。 cout << N