Question

我必须重写fucntion p1（）作为p2（）以完全模仿p1（），仅使用<iomanip>并且我不断收到状态错误

类型为long long unsigned int且“未解析的函数类型”的无效操作数为二进制operator<<

代码在这里：

void p1()
{
    printf("Size of different basic C++ data type in number of bytes\n\n") ;
    printf("size of int           = %d \n", sizeof (int) ) ;
    printf("size of long          = %d \n", sizeof (long) ) ;
    printf("size of short         = %d \n", sizeof (short) ) ;
    printf("size of unsigned int  = %d \n", sizeof (unsigned int) ) ;
    printf("size of char          = %d \n", sizeof (char) ) ;
    printf("size of wchar_t       = %d \n", sizeof (wchar_t) ) ;
    printf("size of bool          = %d \n", sizeof (bool) ) ;
    printf("size of float         = %d \n", sizeof (float) ) ;
    printf("size of double        = %d \n", sizeof (double) ) ;
    printf("size of long double   = %d \n", sizeof (long double) ) ;
    printf("size of int ptr       = %d \n", sizeof (int *) ) ;
    printf("size of double ptr    = %d \n", sizeof (double *) ) ;
    printf("size of char ptr      = %d \n", sizeof (char *) ) ;
    printf("====================================\n\n") ;
}

这是p2（），我必须使用<iomanip>：

void p2()
{
    cout<<"Size of different basic C++ data type in number of bytes\n\n";  
    cout<<setw(10)<<"size of int"<<"= %d" ,sizeof (int)<<endl;
    cout<<setw(10)<<"size of long"<<"= %d" ,sizeof (long)<<endl;
    cout<<setw(10)<<"size of short"<<"= %d", sizeof (short)<<endl ;
    cout<<setw(10)<<"size of unsigned int"<<"= %d", sizeof (unsigned int)<<endl;
    cout<<setw(10)<<"size of char"<<"= %d", sizeof (char)<<endl;
    cout<<setw(10)<<"size of wchar_t"<<"= %d", sizeof (wchar_t)<<endl;
    cout<<setw(10)<<"size of bool"<<"= %d", sizeof (bool)<<endl;
    cout<<setw(10)<<"size of float"<<"= %d", sizeof (float)<<endl;
    cout<<setw(10)<<"size of double"<<"= %d", sizeof (double)<<endl;
    cout<<setw(10)<<"size of long double"<<"= %d", sizeof (long double)<<endl;
    cout<<setw(10)<<"size of int ptr"<<"= %d", sizeof (int *)<<endl;
    cout<<setw(10)<<"size of double ptr"<<"= %d", sizeof (double *)<<endl;
    cout<<setw(10)<<"size of char ptr"<<"= %d", sizeof (char *)<<endl;
    cout<<setfill('=')<<setw(40)<<"="<<endl;
}

Answer 1

声明

cout<<setw(10)<<"size of int"<<"= %d" ,sizeof (int)<<endl;

相当于

( (cout<<setw(10)<<"size of int"<<"= %d") , sizeof (int) ) << endl;

注意那里的逗号 - 它充当comma operator，丢弃其左手参数并返回正确的参数。你得到的基本上是：

sizeof(int) << endl;

显然无效。此外，<< "= %d"只会在流中插入字符串，而不是其他内容。格式说明符在这里没有意义。当您使用operator<<进行流式传输时，重载决策会自动选择正确的重载，以便正确格式化流式传输的值。

你需要

cout << setw(10) << "size of int = " << sizeof(int) << endl;

在Bloodshed中使用<iomanip>使用C ++进行格式化</iomanip>

1 个答案: