使用 cout 将数组和字符打印到屏幕时出现意外结果

Unexpected results when printing array and char to screen with cout(使用 cout 将数组和字符打印到屏幕时出现意外结果)

问题描述

作为一个学习 C++ 的初学者，我试图理解 char 类型的数组和 int 类型的数组之间的区别.这是我的代码:

void IntArray () {
    int array[5] = {5,6,7,8,9};

    cout << "Print int array: " << array << endl;
    cout << "Print int array[0]: " << array[0] << endl;
    cout << "Print int array[0]+1: " << array[0]+1 << endl;
}

void CharArray () {
    char array[5] = {'a', 'b', 'c', 'd', ''};

    cout << "Print char array: " << array << endl;
    cout << "Print char array[0]: " << array[0] << endl;
    cout << "Print char array[0]+1: " << array[0]+1 << endl;
}

Print int array: 0xbfd66a88
Print int array[0]: 5
Print int array[0]+1: 6
Print char array: abcd
Print char array[0]: a
Print char array[0]+1: 98

cout << "Print char array: " << array << endl;

cout << "Print char array[0]+1: " << array[0]+1 << endl;

因为当您将 char 数组流式传输到 cout 时，它们的处理方式与其他数组不同 - << 运算符为 <代码>const char*.这是为了与 C 兼容，因此以 null 结尾的 char 数组被视为字符串.

1.

Because char arrays are treated differently to other arrays when you stream them to cout - the << operator is overloaded for const char*. This is for compatibility with C, so that null-terminated char arrays are treated as strings.

请参阅这个问题.

这是由于整体推广.当您使用 char(值为 'a')和 int(值为 1)调用二进制 + 时，编译器会提升您的char 为 signed int 或 unsigned int.哪一个是特定于实现的——它取决于默认情况下 char 是有符号还是无符号，以及哪个 int 可以采用 char 的全部范围.因此，+ 操作符被调用时使用值 '97' 和 '1'，它返回值 '98'.要将其打印为 char，您需要先转换它:

This is due to integral promotion. When you call the binary + with a char (with value 'a') and an int (with value 1), the compiler promotes your char to either a signed int or an unsigned int. Which one is implementation specific - it depends on whether char is signed or unsigned by default, and which int can take the full range of char. So, the + operator is called with the values '97' and '1', and it returns the value '98'. To print that as a char, you need to first cast it:

cout << "Print char array[0]+1: " << static_cast<char>(array[0]+1) << endl;

请参阅这个问题.

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