Integer Vector element comparison using subscript operator fails, but succeeds using at function

Question

I am writing a callback function which has a reference to a int vector passed to it in a structure. When I try to access an element in the vector using subscript operator[], Intellisense indicates that the == cannot compare the two elements specifically the error is error C2678: binary '==' : no operator found which takes a left-hand operand of type 'std::vector<_Ty>' (or there is no acceptable conversion). But when using at() function there is no problem.

//body of call back function
searchInfo* argVal = (searchInfo*) Parameter;
for(int i = argVal->inclStartPos; i < argVal->exclEndPos; ++i){
    if(argVal->numVector[i] == argVal->searchNum)//problem here
        argVal->result = true;
//this is the structure passed through pointer

struct searchInfo{
int inclStartPos;
int exclEndPos;
vector<int>* numVector; 
int searchNum;
bool result;
};

Since the [] operator and at() function of a vector work almost(the difference doesn't matter here) in the same manner, why the error?

Igor Chornous · Accepted Answer · 2012-02-29 12:28:33Z

6

Actually, you are comparing vector < int > with int, since the field numVector has type

vector<int>*

Simply speaking, you've declared an array of vector's. Operator [] will return a value of type vector.

Probably you have misspelled the declaration. Maybe you wanted to declare numVector as follows:

vector<int> numVector;

edited Feb 29, 2012 at 12:28

answered Feb 29, 2012 at 12:23

Igor Chornous

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3 Comments

Chowlett Over a year ago

Not sure why this was downvoted. It may not be expressed as clearly as possible, but it's right - numVector is a pointer-to-vector, which is compatible with array-of-vector. Hence numVector[i] is a vector<int>; but numVector->at(i) would work.

AquaAsh Over a year ago

@Chowlett: Could you please elaborate on why would numVector->at(i) would work. AFAIK the '==' operator has same data types on both sides , so the expression should work but it doesn't.

Chowlett Over a year ago

@AquaAsh - numVector is a pointer, not an actual vector. Hence numVector[i] will be interpreted as subscripting into an array vector<int>[] from the address numVector, not as taking an element from a vector<int>. numVector[i] == searchNum compares vector<int> with int.numVector->at(i) is calling at on the object pointed to by numVector - it's equivalent to *numVector.at(i)

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