我想實現strassen的算法,我雖然要實現一個矩陣類。但是,我得到以下代碼的段錯誤:分割錯誤實現矩陣類C++
class Matrix {
public:
vector < vector <int> > m;
int r;
int c;
Matrix() {;}
Matrix(int _r, int _c) {
r = _r;
c = _c;
m.resize(r);
for (int i = 0; i < r; i++) {
m[i].resize(c);
}
for(int i =0; i < r; i++){
for(int j=0; j< c; j++){
m[i][j] = 0;
}
}
}
friend ostream& operator<<(ostream &out, Matrix &A) {
for(int i =0; i<A.r; i++){
out << endl;
for(int j=0; j<A.c; j++){
out << A.m[i][j] << "\t";
}
}
out<< endl;
return out;
}
Matrix(const Matrix &A) {
c = A.c;
r = A.r;
m.resize(r);
for (int i = 0; i < r; i++) {
m[i].resize(c);
}
for(int i =0; i<r; i++){
for(int j=0; j<c; j++){
m[i][j] = A.m[i][j];
}
}
}
Matrix& operator-= (Matrix &A) {
assert(A.r == r);
assert(A.c == c);
for(int i =0; i<r; i++){
for(int j=0; j<c; j++){
m[i][j] -= A.m[i][j];
}
}
return *this;
}
Matrix& operator- (Matrix &A) {
Matrix C(*this);
return C-=A;
}
Matrix& operator+= (Matrix &A) {
assert(A.r == r);
assert(A.c == c);
for(int i =0; i<r; i++){
for(int j=0; j<c; j++){
m[i][j] += A.m[i][j];
}
}
return *this;
}
Matrix& operator+ (Matrix &A) {
Matrix C (*this);
(C)+=A;
return C;
}
Matrix getBlock(int sR, int eR, int sC, int eC) {
assert(sR > eR);
assert(sC > eC);
Matrix C(eR-sR, eC-sC);
for(int i = 0; i < C.r; i++) {
for(int j=0; j < C.c; j++) {
C.m[i][j] = m[sR+i][sC+j];
}
}
return C;
}
friend void swap(Matrix& first, Matrix& second) {
using std::swap;
swap(first.r, second.r);
swap(first.c, second.c);
swap(first.m, second.m);
}
Matrix& operator=(Matrix other){
return *this;
}
friend Matrix& operator*(const Matrix& A, const Matrix &B) {
assert(A.r == B.c);
Matrix C(A.r, B.c);
for(int i =0; i<C.r; i++){
for(int j=0; j<C.c; j++){
for(int k = 0; k < A.r; k++) {
C.m[i][j] += A.m[i][k] * B.m[k][j];
}
}
}
return C;
}
};
int main (void)
{
Matrix A(2,2), B(2,2);
A.m[0][0] = 1; A.m[0][1] = 2;
A.m[1][0] = 3; A.m[1][1] = 4;
B.m[0][0] = 1; B.m[0][1] = 2;
B.m[1][0] = 3; B.m[1][1] = 4;
Matrix C(2,2);
C =A+B;
cout << C << endl;
return 0;
}
如果我嘗試A + = B;它的工作原理......我不明白它與A + B有什麼不同。我試圖打印C之前從
Matrix& operator+ (Matrix &
Matrix C (*this);
(C)+=A;
return C;
}
這是正確的。當代碼碰到返回時,我的程序崩潰。我想了解我做錯了什麼。非常感謝。 Davide
將'operator +'的返回類型從'Matrix&'更改爲'Matrix',然後考慮您返回的引用變量的生命週期。 – WhozCraig
[可以訪問局部變量的內存是否可以在其作用域之外訪問?](http://stackoverflow.com/questions/6441218/can-a-local-variables-memory-be-accessed-outside-its-scope) – chris
考慮替換'assert(Ar == Bc);'拋出異常;這使得調用者可以選擇捕捉和繼續,而不是讓程序中止。 –