CODEWITHSUNDEEP
c++vectorpush-back
Simplicity
Simplicity

Reputation: 48916

C++ vector - push_back

In the C++ Primer book, Chapter (3), there is the following for-loop that resets the elements in the vector to zero.

vector<int> ivec; //UPDATE: vector declaration
for (vector<int>::size_type ix = 0; ix ! = ivec.size(); ++ix)
ivec[ix] = 0;

Is the for-loop really assigning 0 values to the elements, or do we have to use the push_back function?

So, is the following valid?

ivec[ix] = ix;

Thanks.

Upvotes: 1

Views: 3234

Answers (5)

Sarfaraz Nawaz
Sarfaraz Nawaz

Reputation: 361264

Is the for-loop really assigning 0 values to the elements? Or, we have to use the push_back finction?

ivec[ix] =0 updates the value of existing element in the vector, while push_back function adds new element to the vector!

So, is the following valid?
ivec[ix] = ix;

It is perfectly valid IF ix < ivec.size().

It would be even better if you use iterator, instead of index. Like this,

int ix = 0;
for(vector<int>::iterator it = ivec.begin() ; it != ivec.end(); ++it)
{ 
     *it = ix++; //or do whatever you want to do with "it" here!
}

Use of iterator with STL is idiomatic. Prefer iterator over index!

Upvotes: 5

CashCow
CashCow

Reputation: 31425

Yes, assuming ix is a valid index, most likely: you have a vector of int though and the index is size_type. Of course you may want to purposely store -1 sometimes to show an invalid index so the conversion of unsigned to signed would be appropriate but then I would suggest using a static_cast.

Doing what you are doing (setting each value in the vector to its index) is a way to create indexes of other collections. You then rearrange your vector sorting based on a predicte of the other collection.

Assuming that you never overflow (highly unlikely if your system is 32 bits or more) your conversion should work.

Upvotes: 0

Varun Madiath
Varun Madiath

Reputation: 3252

The purpose of this for loop is to iterate through the elements of the vector. Starting at element x (when ix is 0) up to the last element (when ix is ivec.size() -1).

On each iteration the current element of the vector is set to 9. This is what the statement

ivec[ix] = 0;

does. Putting

ivec[ix] = ix;

in the for loop would set all the elements of the vector to their position in the vector. i.e, the first element would have a value of zero (as vectors start indexing from 0), the second element would have a value of 1, and so on and so forth.

Upvotes: 0

Mario
Mario

Reputation: 36477

Using the array brackets the vector object acts just like any other simple array. push_back() increases its length by one element and sets the new/last one to your passed value.

Upvotes: 0

templatetypedef
templatetypedef

Reputation: 372664

Yes, you can use the square brackets to retrieve and overwrite existing elements of a vector. Note, however, that you cannot use the square brackets to insert a new element into a vector, and in fact indexing past the end of a vector leads to undefined behavior, often crashing the program outright.

To grow the vector, you can use the push_back, insert, resize, or assign functions.

Upvotes: 2

Related Questions