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我想測試下面的代碼,但是出現編譯錯誤。讓我感到困惑的是,創建和打印pd1和pd2的方式與pd3和pd4相同,但編譯器在打印時會抱怨pd3和pd4。無法打印緩衝區地址
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;
}
編譯錯誤:
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;
在這兩行中用'<<'替換'<'。 – Tobias 2015-04-04 20:27:25