C ++持久性矢量，用文本文件中的数据填充矢量

Question

i am currently trying to learn some C++ and now i got stuck in an exercise with vectors. So the task is to read ints from a text file and store them in the vector which should be dynamic.

I guess there is something wrong with the while-loop? If I start this, the program fails and if I set the vector size to 6, I get 6 0 0 0 0 0 as output.

Thanks for any hints.

int main()
{
const string filename = "test.txt";
int s = 0;
fstream f;
f.open(filename, ios::in);
vector<int> v;

if (f){
    while(f >> s){
        int i = 0;
        v[i] = s;
        i = i+1;
    }

  f.close();
}

for(int i = 0; i < 6; i++){
    cout << v[i] << "\n";
}

}

Answer 1

1. You don't grow the vector. It is empty and cannot hold any int s. You'll need to either resize it every time you want to add another int or you use push_back which automatically enlarges the vector.

2. You set i = 0 for every iteration so you would change the first value of the vector every iteration instead of the next one.

Go for:

    v.push_back(s);

in your loop and

for(int i = 0; i < v.size(); i++) { // ...

Remark:

You normally don't hardcode vector sizes/bounds. One major point about using std::vector is its ability to behave dynamically with respect to its size. Thus, the code dealing with vectors should not impose any restrictions about the size of the vector onto the respective object.

Example:

for(int i = 0; i < 6; i++){ cout << v[i] << "\\n"; }

requires the vector to have at least 6 elements, otherwise (less than 6 ints) you access values out of bounds (and you potentially miss elements if v contains more than 6 values).

Use either

for(int i = 0; i < v.size(); i++){ cout << v[i] << "\\n"; }

or

for(std::vector<int>::const_iterator i = v.begin(); i != v.end(); ++i) { cout << *i << "\\n"; }

or

for(auto i = v.begin(); i != v.end(); ++i) { cout << *i << "\\n"; }

or

for(int x : v){ cout << x << "\\n"; }

or

for(auto && x : v){ cout << x << "\\n"; }

or

std::for_each(v.begin(), v.end(), [](int x){ std::cout << x << "\\n"; });

or variants of the above which possibly pre-store v.size() or v.end()

or whatever you like as long as you don't impose any restriction on the dynamic size of your vector.

Answer 2

Error is this line int i = 0; because you declare i=0 every time in while-loop. To correct this move this line outside from loop.

Note: this will work, if you declare v like normal array for example int v[101] When you use std vectors you can just push element at the end of vector with v.push_back(element);

Answer 3

The issue is in the line i= 0 . Fixing that will give an issue in the line v[i] = s .

You always initialise i to 0 in the while loop, and that is responsible for the current output. You should shift it out of the while loop.

After fixing that, you have not allocated memory to that vector, and so v[i] doesn't make sense as it would access memory beyond bounds. This will give a segmentation fault. Instead, it should be v.push_back(i) , as that adds elements to the end of a vector, and also allocates memory if needed.

Answer 4

如果您使用的是std::vector ，则可以使用v.push_back(i)填充此向量

Answer 5

v[i] = s; //error,you dont malloc room for vector

更改为： v.push_back(s);