请问为什么结果不同?
C++代码:
MatrixXcd testTest;
testTest.resize(3,3);
testTest.real()(0,0) = 1;
testTest.real()(0,1) = 2;
testTest.real()(0,2) = 3;
testTest.real()(1,0) = 1;
testTest.real()(1,1) = 2;
testTest.real()(1,2) = 3;
testTest.real()(2,0) = 1;
testTest.real()(2,1) = 2;
testTest.real()(2,2) = 3;
testTest.imag()(0,0) = 1;
testTest.imag()(0,1) = 2;
testTest.imag()(0,2) = 3;
testTest.imag()(1,0) = 1;
testTest.imag()(1,1) = 2;
testTest.imag()(1,2) = 3;
testTest.imag()(2,0) = 1;
testTest.imag()(2,1) = 2;
testTest.imag()(2,2) = 3;
cout<< endl << testTest << endl;
cout<< endl << testTest.transpose() << endl;
cout<< endl << testTest*testTest.transpose() << endl;
cout<< endl << testTest << endl;
C++的结果:
(1,1) (2,2) (3,3)
(1,1) (2,2) (3,3)
(1,1) (2,2) (3,3)
(1,1) (1,1) (1,1)
(2,2) (2,2) (2,2)
(3,3) (3,3) (3,3)
(0,28) (0,28) (0,28)
(0,28) (0,28) (0,28)
(0,28) (0,28) (0,28)
(1,1) (2,2) (3,3)
(1,1) (2,2) (3,3)
(1,1) (2,2) (3,3)
同样的内容用Matlab写成:
testTest = [ complex(1,1) complex(2,2) complex(3,3);
complex(1,1) complex(2,2) complex(3,3);
complex(1,1) complex(2,2) complex(3,3)];
testTest
testTest'
testTest*testTest'
testTest
Matlab结果:
testTest =
1.0000 + 1.0000i 2.0000 + 2.0000i 3.0000 + 3.0000i
1.0000 + 1.0000i 2.0000 + 2.0000i 3.0000 + 3.0000i
1.0000 + 1.0000i 2.0000 + 2.0000i 3.0000 + 3.0000i
ans =
1.0000 - 1.0000i 1.0000 - 1.0000i 1.0000 - 1.0000i
2.0000 - 2.0000i 2.0000 - 2.0000i 2.0000 - 2.0000i
3.0000 - 3.0000i 3.0000 - 3.0000i 3.0000 - 3.0000i
ans =
28 28 28
28 28 28
28 28 28
testTest =
1.0000 + 1.0000i 2.0000 + 2.0000i 3.0000 + 3.0000i
1.0000 + 1.0000i 2.0000 + 2.0000i 3.0000 + 3.0000i
1.0000 + 1.0000i 2.0000 + 2.0000i 3.0000 + 3.0000i
在C中,testTest * testTest'的乘积返回具有实部为0和虚部为28的复数。而在Matlab中,返回值只是一个值为28的double类型变量。
.conjugate()
,但如果你使用.adjoint()
,它会一次完成两个操作。 - DavidW