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C++ ';表';尝试用C+;打印数组时格式不好+;_C++ - Fatal编程技术网

C++ ';表';尝试用C+;打印数组时格式不好+;

C++ ';表';尝试用C+;打印数组时格式不好+;,c++,C++,因此,我试图打印出一张表格,并均匀填充,如下所示: int rowLength = 8 * 4 + 8 + 1; char arr[rowLength]; fill(arr, arr + rowLength, '-'); string divider = arr; for(int y = 0; y < 5; y++) { cout << "\n" << divider << "\n"; for(int x = 0; x < 8;

因此,我试图打印出一张表格,并均匀填充,如下所示:

int rowLength = 8 * 4 + 8 + 1;
char arr[rowLength];

fill(arr, arr + rowLength, '-');
string divider = arr;

for(int y = 0; y < 5; y++) {
    cout << "\n" << divider << "\n";
    for(int x = 0; x < 8; x++) {
        cout << "|  " << "1";
        if(x == 7)
            cout << "|";
    }
}
cout << "\n" << divider;
int rowLength=8*4+8+1;
字符arr[行长度];
填充(arr,arr+行长“-”);
字符串分隔符=arr;
对于(int y=0;y<5;y++){
库特
此
std::string
arr
构造而成,arr
是一个
char
数组,该数组衰减为
char*
。从
char*
构造时,
char*
必须是以空结尾的字符串

您的代码用字符填充整个
char
数组,但无法追加额外的
'\0'
字节。这将导致未定义的行为

std::string
arr
构造而成,arr是一个
char
数组,该数组衰减为
char*
。从
char*
构造时,
char*
必须是以空结尾的字符串


您的代码用字符填充整个
char
数组,但无法追加额外的
'\0'
字节。这将导致未定义的行为。

Edited:added divider code Edited:added divider code
string divider = arr;