Question

我想测试以下代码，但是我收到了编译错误。令我困惑的是，创建和打印pd1和pd2的方式与pd3和pd4相同，但编辑器在打印时会抱怨pd3和pd4。

const int BUF = 512;
const int N = 5;
char buffer[BUF];       // chunk of memory
int main(){
    using namespace std;

    double *pd1, *pd2;
    int i;
    cout << "Calling new and placement new:\n";
    pd1 = new double[N];            // use heap
    pd2 = new (buffer) double[N];   // use buffer array
    for (i = 0; i < N; i++) 
        pd2[i] = pd1[i] = 1000 + 20.0*i;
    cout << "Buffer addresses:\n" << " heap: " << pd1 << " static: " << (void *)buffer << endl;
    cout << "Buffer contents:\n";
    for(i = 0; i < N; i++) {
        cout << pd1[i] << " at " << &pd1[i] << "; ";
        cout << pd2[i] << " at " << &pd2[i] << endl;
    }

    cout << "\nCalling new and placement new a second time:\n";
    double *pd3, *pd4;
    pd3 = new double[N];
    pd4 = new (buffer) double[N];
    for(i = 0; i < N; i++) 
        pd4[i] = pd3[i] = 1000 + 20.0 * i;
    cout << "Buffer contents:\n";
    for (i = 0; i < N; i++) {
        cout << pd3[i] < " at " << &pd3[i] << "; ";
        cout << pd4[i] < " at " << &pd4[i] << endl;
    }

    return 0;
}

编译错误：

newplace.cpp: In function ‘int main()’:
newplace.cpp:33:36: error: invalid operands of types ‘const char [5]’ and ‘double*’ to binary ‘operator<<’
   cout << pd3[i] < " at " << &pd3[i] << "; ";
                                    ^
newplace.cpp:34:36: error: invalid operands of types ‘const char [5]’ and ‘double*’ to binary ‘operator<<’
   cout << pd4[i] < " at " << &pd4[i] << endl;

Answer 1

你错过了一个＆lt;符号在这里

 cout << pd3[i] < " at " << &pd3[i] << "; ";
 cout << pd4[i] < " at " << &pd4[i] << endl;

尝试

 cout << pd3[i] << " at " << &pd3[i] << "; ";
 cout << pd4[i] << " at " << &pd4[i] << endl;

Answer 2

您只在要尝试打印缓冲区内容的流中放置一个<。

cout << pd3[i] < " at " << &pd3[i] << "; "; // there is only one <
cout << pd4[i] < " at " << &pd4[i] << endl; // ^

确保您在流插入运算符中有两个<'s。

cout << pd3[i] << " at " << &pd3[i] << "; ";
cout << pd4[i] << " at " << &pd4[i] << endl;

无法打印缓冲区地址

2 个答案: