enum Direction {
NORTH = 0,
EAST = 1,
SOUTH = 2,
WEST = 3 };
struct Point {
int x, y; };
class Path {
private:
Point visited[10000];
int visited_positions;
bool circular;
public:
Path() {
visited_positions = 1;
circular = 0;
Point p;
p.x = 0;
p.y = 0;
visited[0] = p;
}
void add_point(Point p) {
if(!circular) {
for(int i = 0; i < visited_positions; i++) {
if(visited[i].x == p.x && visited[i].y == p.y) {
circular = true;
break;
}
}
}
visited[visited_positions] = p;
visited_positions++;
}
bool is_circular() {
return circular;
}
};
class Robot {
private:
Point position;
Direction direction;
Path path;
public:
Robot() {
position.x = 0;
position.y = 0;
direction = NORTH;
}
void turn_left() {
direction = static_cast<Direction>((direction + 3) % 4);
}
void turn_right() {
direction = static_cast<Direction>((direction + 1) % 4);
}
void forward() {
switch (direction) {
case NORTH:
position.y++;
break;
case EAST:
position.x++;
break;
case SOUTH:
position.y--;
break;
case WEST:
position.x--;
break;
}
path.add_point(position);
}
Point get_position() {
return position;
}
bool has_gone_in_a_loop() {
return path.is_circular();
} };
int main() {
int test_cases;
cin >> test_cases;
string instruction_string;
for(int tc = 0; tc != test_cases; tc++) {
cin >> instruction_string;
Robot skundi;
for(size_t i = 0; i < instruction_string.size(); i++) {
switch(instruction_string[i]) {
case 'H':
skundi.turn_right();
break;
case 'V':
skundi.turn_left();
break;
case 'F':
skundi.forward();
break;
}
}
if(skundi.has_gone_in_a_loop()) {
cout << "Circular" << endl;
}
else {
cout << "OK" << endl;
}
}
return 0; }
这是一项家庭作业,我会很感激任何提示。 (请不要赠品:D) 这些是我得到的valgrind错误。
Memcheck, a memory error detector
Invalid read of size 8
Using Valgrind:
**HEAP SUMMARY:**
in use at exit: 16,352 bytes in 1 blocks
total heap usage: 14 allocs, 13 frees, 28,907 bytes allocated
16,352 bytes in 1 blocks are possibly lost in loss record 1 of 1
at 0xX: operator new(unsigned long) (vg_replace_malloc.c:298....)
by 0xX: std::string::_Rep::_S_create(unsigned long, unsigned long, std::allocator const&) (in /usr/lib64/libstdc++.so.6.0.13)
by 0xX: std::string::_Rep::_M_clone(std::allocator const&, unsigned long) (in /usr/lib64/libstdc++.so.6.0.13)
by 0xX: std::string::reserve(unsigned long) (in /usr/lib64/libstdc++)
by 0xX: std::basic_istream >& std::operator>>, std::allocator >(std::basic_istream >&, std::basic_string, std::allocator >&) (in /usr/lib64/libstdc++)
**LEAK SUMMARY:**
definitely lost: 0 bytes in 0 blocks
indirectly lost: 0 bytes in 0 blocks
possibly lost: 16,355 bytes in 1 blocks
still reachable: 0 bytes in 0 blocks
suppressed: 0 bytes in 0 blocks
For counts of detected and suppressed errors, rerun with: -v
ERROR SUMMARY: 1 errors from 1 contexts (suppressed: 6 from 6)
答案 0 :(得分:1)
这不是严格意义上的内存泄漏,但您可以将其视为一个(这就是为什么它看起来“可能丢失”)。
这是复制问题的最小方案:
static int *global_data = nullptr;
int main(int, char**)
{
global_data = new int{9}; // allocated for the duration of the application
// we rely on the program exitting for deallocation
}
在这种情况下,内存永远不会丢失(因为你仍然有一个指向它的指针)。它也永远不会被释放,因此数据是否泄漏是值得商榷的(在各种论坛上反复讨论的事情)。
这就是为什么valgrind在这个问题上占据中间地位并说“可能会丢失”(由你来决定这是否是泄漏,取决于你想要做什么)。
在您的具体示例中,“可能丢失”的内容是stl实现的内部数据。没有什么可担心的。